- Determine whether the constant derivative is positive, negative, or zero. - A derivative that never changes value imposes a very specific shape on the original graph. - Use the sign of the derivative to decide which direction the graph must move.

1. Since \(3>1\), \(\ln(3)>0\). Thus \(f'\) is a positive constant. 2. A constant positive derivative means \(f\) is a line with constant positive slope. 3. Only graph a has that form, so graph a could represent \(f\).

Graph a

- The sign of \(f'\) gives the direction of travel. - A negative-to-positive change in \(f'\) gives a local minimum. - The sign of \(f''\) gives concavity. - Use \(f(0)=1\) to place the landmark vertically.

1. Since \(f'(x)<0\) for \(x<0\), the graph decreases as it approaches \(x=0\) from the left. 2. Since \(f'(x)>0\) for \(x>0\), the graph increases to the right of \(x=0\). Thus \(f'\) changes from negative to positive, so \(x=0\) is a local minimum. 3. The anchor \(f(0)=1\) makes the minimum point \((0,1)\). 4. Because \(f''(x)>0\) everywhere, both the left and right branches are concave up. Blueprint: decreasing and concave up to \((0,1)\), then increasing and concave up.

Decreasing and concave up for \(x<0\); local minimum at \((0,1)\); increasing and concave up for \(x>0\).

- Above the x-axis, the derivative has what sign? - Translate a derivative sign into the direction of the original graph. - When the sign changes at a zero, what shape does the original graph have nearby?

1. The derivative \(f'(x)=2-x\) is positive for \(x<2\) and negative for \(x>2\). 2. Therefore, a graph of \(f\) rises up to \(x=2\) and falls after \(x=2\). 3. The graph has a local maximum at \(x=2\). Its vertical position is not determined by \(f'\).

A graph of \(f\) rises for \(x<2\), has a local maximum at \(x=2\), and falls for \(x>2\). The height of the graph cannot be determined from \(f'\) alone.

- What does a horizontal derivative graph say about the slope of the original graph? - Does a constant slope make a curve or a straight line? - Could two vertically shifted straight lines share the same derivative?

1. The derivative is the constant \(f'(x)=-2\), so the slope of \(f\) is always \(-2\). 2. A function with this derivative has the form \(f(x)=-2x+C\), where \(C\) is any constant. 3. Thus its graph is a straight line sloping downward. The derivative graph does not determine \(C\) or the line's height.

The graph of \(f\) is a decreasing straight line with slope \(-2\). Its height is not determined.

- What does the sign of the second derivative tell you about tangent slopes? - Curving upward and moving upward are not the same thing. - Think of a U-shaped curve: how can it move on its left and right sides?

1. The graph shows \(f''(x)>0\) throughout \((-2,2)\). 2. Thus the slope \(f'(x)\) increases, and the graph of \(f\) is concave up. 3. Without knowing the sign of \(f'\), \(f\) may be increasing, decreasing, or change from decreasing to increasing.

The graph of \(f\) is concave up, and its tangent slopes increase. Whether \(f\) increases or decreases cannot be determined from \(f''\) alone.

- Think about how tangent slopes change as you move from left to right. - Which curve becomes less steep upward, then more steep downward? - Do not decide based only on whether a curve is rising or falling.

1. Negative second derivative means the tangent slopes decrease as x increases. 2. Curve a) opens downward and has decreasing tangent slopes. Curve b) opens upward and has increasing tangent slopes. 3. Therefore, curve a) fits the condition.

Curve a), because it bends downward and its tangent slopes decrease from left to right.

- Interpret the first derivative before thinking about curvature. - What does a negative second derivative tell you about how slopes change? - Can a rising graph get flatter?

1. Since \(f'(x)>0\), the graph rises throughout the interval. 2. Since \(f''(x)<0\), its tangent slopes decrease; the graph bends downward. 3. Thus the graph rises but becomes less steep as x increases.

The graph rises and is concave down. Its positive slopes become smaller from left to right.

- Which graphs have the tangent slopes of an upward-opening parabola? - Look at where each graph meets the y-axis. - A derivative determines shape but does not fix a vertical shift on its own.

1. A function with derivative \(2x\) is a vertical shift of \(x^2\), so it opens upward. 2. The point condition fixes its height: the graph must pass through \((0,3)\). 3. Graph a) opens upward and passes through \((0,3)\), so it is the correct choice.

Graph a).

- Find where the derivative changes sign. - Turn the derivative signs into a rise-fall pattern. - Which graph has a peak first and a valley second?

1. Factor \(f'(x)=(x-1)(x+1)\). It is positive for \(x<-1\) and \(x>1\), and negative for \(-1<x<1\). 2. A compatible function rises, then falls, then rises, with a local maximum at \(-1\) and a local minimum at \(1\). 3. Graph a) has this sequence. Graph b) falls, rises, then falls.

Graph a).

- How are the critical numbers of a function related to the zeros of its derivative? - On which intervals is each graph increasing or decreasing? - What sign should the derivative have when the original function is increasing or decreasing? - Check both possible assignments.

1. The graph labeled \(f\) has a local maximum between \(x=-2\) and \(x=-1\), and a local minimum between \(x=1\) and \(x=2\). The graph labeled \(g\) is zero at those same x-coordinates. 2. The graph of \(f\) increases before the first critical number, decreases between the two critical numbers, and increases after the second critical number. Over those intervals, \(g\) is positive, negative, and positive, respectively. 3. Therefore, the values of \(g\) match the slopes of \(f\), so \(g=f'\).

The graph labeled \(f\) represents the function, and the graph labeled \(g\) represents its derivative. Thus, \(g=f'\).

- The derivative represents the slope of the graph at each input. - Relate increasing and decreasing behavior to the sign of the derivative. - At a smooth local maximum or minimum, the tangent is horizontal. - Compare the derivative's sign pattern with the opening direction of a parabola.

1. A is false. At \(x=0\), the graph of \(f\) is decreasing, so \(f'(0)<0\). Passing through the origin does not determine the derivative sign. 2. B is true. The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\), so both tangent lines are horizontal. 3. C is true. The function decreases throughout \((-2,2)\), so \(f'(x)<0\) on that interval. 4. D is false. The derivative is positive outside the two zeros and negative between them, which matches an upward-opening parabola.

A) False; \(f'(0)<0\) because \(f\) is decreasing there. B) True; both inputs are smooth extrema. C) True; \(f\) is decreasing on \((-2,2)\). D) False; the sign pattern matches an upward-opening parabola.

- First decide which graph could be the derivative by comparing zeros on one graph with horizontal tangents on the other. - Only after identifying the roles should you read function values and derivative values. - A requested tangent slope is a requested value of \(f'\).

1. Figure 1 is \(f\) and Figure 2 is \(f'\). The turning points of Figure 1 occur at the same inputs where Figure 2 crosses the x-axis, which is the required function-derivative relationship. 2. From Figure 1, \(f(2)=-1\). 3. The slope at \(x=-1\) is \(f'(-1)\). Figure 2 gives \(f'(-1)=-1.5\). 4. Figure 1 passes through \((0,1)\), so \(a=0\) works. 5. A slope of \(3\) means \(f'(b)=3\). Figure 2 has value \(3\) at \(b=-2\) and \(b=2\).

a) Figure 1 is \(f\); Figure 2 is \(f'\). b) \(f(2)=-1\) c) \(-1.5\) d) \(a=0\) e) \(b=-2\) and \(b=2\)

- Compare the turning points of one graph with the zeros of the other. - A differentiable function has derivative \(0\) at a smooth local maximum or minimum. - Compare where one graph increases or decreases with the sign of the other graph. - After identifying the graphs, read each requested value from the correct one.

1. The graph labeled \(k_1\) has a local maximum at \(x=1\) and a local minimum at \(x=3\). The graph labeled \(k_2\) is zero at those same x-values. Also, \(k_1\) decreases on \((1, 3)\), where \(k_2\) is negative. Therefore, \(k_1\) represents \(f\) and \(k_2\) represents \(f'\). 2. Reading the values from the graphs gives \(f(2)=2\), \(f'(2)=-1.5\), \(f(3)=1\), and \(f'(1)=0\).

a) \(k_1\) is the graph of \(f\), and \(k_2\) is the graph of \(f'\). b) \(f(2)=2\); \(f'(2)=-1.5\); \(f(3)=1\); \(f'(1)=0\)

- Use the sign of \(f'\) for monotonicity and local extrema. - Use whether \(f'\) rises or falls for concavity. - Keep those two readings of the derivative graph separate.

1. Zeros and signs of \(f'\) determine critical points and monotonicity of \(f\), while local extrema of \(f'\) can indicate concavity changes of \(f\). 2. At \(x=-2\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x=2\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. Statement a is true. 3. On \((-2,2)\), \(f'(x)>0\), so \(f\) is strictly increasing, not decreasing. Statement b is false. 4. The graph of \(f'\) has a local maximum at \(x=0\), so \(f'\) changes from increasing to decreasing. Therefore \(f\) changes from concave up to concave down at \(x=0\). Statement c is true.

a) True. b) False. c) True.

- Find the inputs where \(f\) has horizontal tangents. - The derivative is zero at those inputs. - Determine where \(f\) is decreasing. - The derivative must be negative on decreasing intervals.

1. The graph of \(f\) has a local maximum at \(x=-2\) and a local minimum at \(x=2\). Therefore, \(f'(-2)=0\) and \(f'(2)=0\). This eliminates graph \(C\), which has only one zero. 2. The graph of \(f\) decreases between \(-2\) and \(2\), so \(f'(x)<0\) on that interval. 3. Graph \(A\) is below the x-axis between its zeros at \(-2\) and \(2\), while graph \(B\) is above the x-axis there. Therefore, graph \(A\) represents \(f'\).

Graph \(A\) represents \(f'\).

- Locate the horizontal tangents of \(f\). - The derivative has zeros at those x-coordinates. - Determine where \(f\) is increasing and decreasing. - Eliminate candidates using both zeros and sign.

1. The graph of \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=3\). Thus, \(f'(-1)=0\) and \(f'(3)=0\). This eliminates \(C\), whose zeros are different, and \(D\), which has only one zero. 2. The function decreases between \(-1\) and \(3\), so \(f'(x)<0\) on that interval. 3. Only graph \(A\) has zeros at \(-1\) and \(3\) and is negative between them. Therefore, graph \(A\) represents \(f'\).

Graph \(A\) represents \(f'\).

- Match the zeros of a derivative with the horizontal tangents of its function. - Compare the sign of each possible derivative with where the function increases or decreases. - Use the critical \(x\)-values as interval endpoints.

1. The graph of \(f_1\) has a local maximum at \(x=-2\) and a local minimum at \(x=2\), so its derivative must be zero at those values. 2. Between the two critical numbers, \(f_1\) decreases, so its derivative must be negative there. Graph \(p\) is negative between its zeros, while graph \(q\) is positive. Therefore, \(f_1'=p\) and \(f_2'=q\). 3. Graph \(p\) is positive on \([-3.5, -2)\) and \((2, 3.5]\), and negative on \((-2, 2)\). Thus, \(f_1\) is strictly increasing on \([-3.5, -2]\) and \([2, 3.5]\), and strictly decreasing on \([-2, 2]\).

a) \(f_1 \rightarrow p\) and \(f_2 \rightarrow q\) b) Increasing on \([-3.5, -2]\) and \([2, 3.5]\); decreasing on \([-2, 2]\).

- Compare where the function increases or decreases with the sign of the derivative. - The zeros of the derivative occur at horizontal tangents of the function. - The value \(f'(0)\) is the slope of \(f\) at \(x = 0\).

1. Figure 1 represents \(f\), and Figure 2 represents \(f'\). Figure 1 increases for \(x < -1\), decreases for \(-1 < x < 3\), and increases for \(x > 3\). Over the same intervals, Figure 2 is positive, negative, and positive. 2. The function \(f\) is strictly decreasing on \([-1, 3]\). The derivative graph is below the \(x\)-axis for \(-1 < x < 3\) and is zero at the endpoints. 3. The tangent slope at \(x = 0\) is \(f'(0)\). From Figure 2, \(f'(0) = -0.75\).

a) Figure 1 shows \(f\), and Figure 2 shows \(f'\). b) \([-1, 3]\); the graph of \(f'\) is below the \(x\)-axis between \(-1\) and \(3\). c) \(-0.75\)

- Look for inputs where the graph of \(f\) has horizontal tangents. - Relate increasing behavior of \(f\) to the sign of \(f'\). - For the estimate at \(x=0\), imagine a tangent line and use two convenient points to form a slope triangle.

1. The graph of \(f\) has a local minimum at \(x=-1\) and a local maximum at \(x=1\). Therefore, the zeros of \(f'\) are \(x=-1\) and \(x=1\). 2. The graph of \(f\) increases between the two extrema, so \(f'(x)>0\) for \(-1<x<1\). 3. At \(x=0\), a reasonable tangent estimate uses \((0,1)\) and approximately \((1,2.5)\), giving \(f'(0)\approx\frac{2.5-1}{1-0}=1.5\). Values from about \(1.3\) to \(1.7\) are consistent with the displayed graph.

a) \(x=-1\) and \(x=1\) b) \((-1,1)\) c) Approximately \(1.5\); estimates from about \(1.3\) to \(1.7\) are acceptable.

- Differentiation lowers the degree of a polynomial by one. - Match local extrema of a function with zeros of its derivative. - Compare where each function increases or decreases with the sign of each derivative graph.

1. The graph of \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=3\). Graph \(p\) is zero at those same values. It is negative between them, where \(f\) decreases, and positive outside them, where \(f\) increases. Therefore, \(f'=p\). 2. The graph of \(g\) has a local maximum at \(x=2\). Graph \(q\) is zero there, positive for \(x<2\), where \(g\) increases, and negative for \(x>2\), where \(g\) decreases. Therefore, \(g'=q\).

a) \(f\rightarrow p\) and \(g\rightarrow q\) b) The zeros and signs of \(p\) match the critical numbers and monotonicity of \(f\), while the zero and signs of \(q\) match those of \(g\).

- Identify where the graph of \(f'\) is increasing and decreasing. - How does an increasing or decreasing first derivative determine concavity? - Inflection-point x-coordinates of \(f\) occur where the monotonicity of \(f'\) changes.

1. The function \(f\) is concave up where \(f'\) is increasing and concave down where \(f'\) is decreasing. 2. On the displayed interval, \(f'\) is increasing on \([-5, -1)\) and \((3, 6]\). Therefore, \(f\) is concave up on those intervals. 3. The graph of \(f'\) is decreasing on \((-1, 3)\). Therefore, \(f\) is concave down on that interval. 4. The monotonicity of \(f'\) changes at its local extrema, \(x=-1\) and \(x=3\). Thus these are the x-coordinates of the inflection points of \(f\).

a) Concave up on \([-5, -1)\) and \((3, 6]\); concave down on \((-1, 3)\). b) \(x=-1\) and \(x=3\)

- How does a point’s position relative to the x-axis determine the sign of \(f(x)\)? - What does the tangent slope tell you about the sign of \(f'(x)\)? - What is special about the tangent at a local maximum or minimum? - How does concavity determine the sign of \(f''(x)\)? - What happens to concavity at an inflection point?

1. Use each point’s position relative to the x-axis to determine the sign of \(f(x)\): \(f(x_A)<0\), \(f(x_B)>0\), \(f(x_C)>0\), and \(f(x_D)<0\). 2. Use the tangent slope to determine the sign of \(f'(x)\). The graph is increasing at \(A\), has a horizontal tangent at the local maximum \(B\), is decreasing at \(C\), and has a horizontal tangent at the local minimum \(D\). Therefore, \(f'(x_A)>0\), \(f'(x_B)=0\), \(f'(x_C)<0\), and \(f'(x_D)=0\). 3. Use concavity to determine the sign of \(f''(x)\). The graph is concave down at \(A\) and \(B\), changes concavity at \(C\), and is concave up at \(D\). Therefore, \(f''(x_A)<0\), \(f''(x_B)<0\), \(f''(x_C)=0\), and \(f''(x_D)>0\).

The completed table is: <table border="1" style="border-collapse: collapse; text-align: center; margin-top: 10px;"> <thead> <tr> <th style="padding: 5px 10px; background-color: #f2f2f2;">Point</th> <th style="padding: 5px 10px; background-color: #f2f2f2;">\(f(x)\)</th> <th style="padding: 5px 10px; background-color: #f2f2f2;">\(f'(x)\)</th> <th style="padding: 5px 10px; background-color: #f2f2f2;">\(f''(x)\)</th> </tr> </thead> <tbody> <tr> <td style="padding: 5px 10px; font-weight: bold;">\(A\)</td> <td style="padding: 5px 10px; color: red;">negative (\(<0\))</td> <td style="padding: 5px 10px; color: green;">positive (\(>0\))</td> <td style="padding: 5px 10px; color: red;">negative (\(<0\))</td> </tr> <tr> <td style="padding: 5px 10px; font-weight: bold;">\(B\)</td> <td style="padding: 5px 10px; color: green;">positive (\(>0\))</td> <td style="padding: 5px 10px;">zero (\(=0\))</td> <td style="padding: 5px 10px; color: red;">negative (\(<0\))</td> </tr> <tr> <td style="padding: 5px 10px; font-weight: bold;">\(C\)</td> <td style="padding: 5px 10px; color: green;">positive (\(>0\))</td> <td style="padding: 5px 10px; color: red;">negative (\(<0\))</td> <td style="padding: 5px 10px;">zero (\(=0\))</td> </tr> <tr> <td style="padding: 5px 10px; font-weight: bold;">\(D\)</td> <td style="padding: 5px 10px; color: red;">negative (\(<0\))</td> <td style="padding: 5px 10px;">zero (\(=0\))</td> <td style="padding: 5px 10px; color: green;">positive (\(>0\))</td> </tr> </tbody> </table>

- How does the sign of the second derivative determine concavity? - An inflection point requires a change in the sign of the second derivative. - What condition must hold for the first derivative to have a local extremum?

1. The graph of \(f''\) is above the x-axis on \([-4, -2)\) and \((2, 4]\). Therefore, \(f\) is concave up on those intervals. It is below the x-axis on \((-2, 2)\), so \(f\) is concave down there. 2. At \(x=-2\) and \(x=2\), the second derivative is zero and changes sign. Therefore, these are the x-coordinates of the inflection points of \(f\). 3. The statement is false. A local extremum of \(f'\) at \(x=0\) would require \(f''(0)=0\). The graph shows \(f''(0)=-4\), so \(f'\) does not have a local extremum there.

a) Concave up on \([-4, -2)\) and \((2, 4]\); concave down on \((-2, 2)\). b) \(x=-2\) and \(x=2\) c) False, because \(f''(0)=-4\ne0\).

- Use the turning points to find inputs where a differentiable function must have derivative zero. - Ask which derivative signs are compatible with the given intervals of increase and decrease. - What does strict increase tell you about an interval, and what must hold at individual points?

1. Because \(f\) has a local maximum at \(-1\) and a local minimum at \(3\), and is differentiable there, \(f'(-1)=f'(3)=0\). 2. Where \(f\) is increasing, its derivative is nonnegative; where it is decreasing, its derivative is nonpositive. Strict increase alone does not require \(f'>0\) at every point. 3. Graph I has the required zeros and a compatible positive-negative-positive sign pattern. Graph II reverses the signs, and graph III has different zeros. 4. Thus, graph I could represent \(f'\).

Graph I. Its zeros at \(x=-1\) and \(x=3\) match the turning points of \(f\), and its signs agree with the given intervals of increase and decrease. II has the opposite signs; III has different zeros.

- Zeros of the derivative correspond to critical numbers of the original function. - The sign of the derivative determines increasing and decreasing intervals. - Check both the critical numbers and the direction of each sign change.

1. The derivative graph has zeros at \(x=1\) and \(x=3\), so \(f\) must have critical numbers at those x-values. 2. Candidate 2 has extrema at \(x=0\) and \(x=2\), so it cannot be \(f\). 3. For \(x<1\), \(f'(x)>0\), so \(f\) must be increasing. Candidate 1 increases before \(x=1\) and has a local maximum there because \(f'\) changes from positive to negative. 4. Candidate 3 decreases before \(x=1\), which contradicts \(f'(x)>0\). Therefore, candidate 1 represents \(f\).

a) Candidate 1. b) Candidate 2 has critical numbers at the wrong x-values, and candidate 3 has the wrong monotonicity.

- Use derivative zeros and sign changes for part a). - Verify the candidate function by differentiating it, not by reversing differentiation. - Substitute the critical x-values into the verified function to get coordinates.

1. The zeros of \(g'\) are \(x=-3\) and \(x=3\). At \(x=-3\), \(g'\) changes from negative to positive, so \(g\) has a local minimum. At \(x=3\), \(g'\) changes from positive to negative, so \(g\) has a local maximum. 2. Differentiating \(g(x)=-\frac{1}{18}x^3+1.5x+1\) gives \(g'(x)=-\frac16x^2+1.5\), which matches the displayed derivative. 3. \(g(-3)=-2\) and \(g(3)=4\). Therefore, the local minimum is \((-3,-2)\), and the local maximum is \((3,4)\).

a) Local minimum at \(x=-3\); local maximum at \(x=3\). b) \(g'(x)=-\frac16x^2+1.5\), matching the graph. c) Local minimum \((-3,-2)\); local maximum \((3,4)\).

- A smooth minimum of \(f\) must occur where \(f'\) is zero. - Identify the roles of the two curves before reading the requested slope. - Equal function and derivative values occur where the two curves intersect.

1. Curve \(p\) is \(f\), and curve \(q\) is \(f'\). Curve \(p\) has a minimum at \(x=1\), while curve \(q\) has value \(0\) there, as a derivative must at a smooth minimum. 2. At \(x=0\), curve \(q\) has value \(-1\), so the tangent slope is \(f'(0)=-1\). 3. The function and derivative have equal values where curves \(p\) and \(q\) intersect. The intersections occur at \(x=1\) and \(x=3\).

a) \(p=f\) and \(q=f'\) b) \(-1\) c) \(x=1\) and \(x=3\)

- At a local maximum or minimum, the first derivative is zero. - Compare where one graph increases or decreases with where a candidate derivative is positive or negative. - Build a two-step derivative chain among the three graphs.

1. Graph b has a local maximum at \(x=0\). Its derivative must be zero there and change from positive to negative. Graph c has exactly that behavior, so graph c is the derivative of graph b. 2. Graph c has a local minimum at \(x=1\) and a local maximum at \(x=3\). Its derivative must have zeros at \(x=1\) and \(x=3\). Graph a has those zeros and its sign matches the increasing/decreasing behavior of graph c. Therefore, graph a is the derivative of graph c. 3. The derivative chain is graph b \(\to\) graph c \(\to\) graph a. Hence graph b is \(f\), graph c is \(f'\), and graph a is \(f''\).

Graph a: \(f''\); graph b: \(f\); graph c: \(f'\).

- What must the derivative equal at a smooth local maximum or minimum? - Compare the \(x\)-coordinate of the vertex of \(f\) with the zeros of \(g\).

1. The graph of \(f\) has a local minimum at \(x=0\). 2. If \(g=f'\), then \(g(0)\) would have to equal \(0\), because the tangent to \(f\) is horizontal at a smooth local minimum. 3. The graph shows \(g(0)=1\), not \(0\). Therefore, \(g\) cannot be the derivative of \(f\).

The function \(f\) has a local minimum at \(x = 0\), so its derivative must be \(0\) there. Since \(g(0) = 1\), \(g\) cannot equal \(f'\).

- What sign must a derivative have where its function is decreasing? - Compare that sign with the graph of \(f\) for \(x>0\).

1. For \(x>0\), the graph of \(g\) is strictly decreasing. 2. If \(f=g'\), then \(f(x)\) would have to be negative wherever \(g\) is decreasing. 3. For \(x>0\), the graph of \(f\) is above the x-axis, so \(f(x)>0\). This contradicts the required sign of the derivative. Therefore, \(f\) cannot be the derivative of \(g\).

For \(x > 0\), \(g\) is decreasing, so its derivative must be negative. Since \(f(x) > 0\) in that region, \(f\) cannot equal \(g'\).

- The derivative is zero where the original graph has a horizontal tangent line. - Determine where \(f\) is increasing and decreasing. - The derivative of a quadratic function is linear.

1. The derivative is zero where the graph of \(f\) has a horizontal tangent line. The vertex occurs at \(x=0\), so \(f^{\prime}(0)=0\). 2. The graph of \(f\) is a downward-opening parabola. It increases for \(x<0\), so \(f^{\prime}(x)>0\), and decreases for \(x>0\), so \(f^{\prime}(x)<0\). From the vertex \((0,2)\) and the point \((2,0)\), write \(f(x)=ax^2+2\) and solve \(0=4a+2\), giving \(a=-0.5\). Thus, \(f(x)=-0.5x^2+2\) and \(f^{\prime}(x)=-x\). The derivative graph is a decreasing line through the origin, crossing the x-axis from positive values to negative values.

1) The zero of \(f^{\prime}\) is \(x=0\). 2) The graph is the line \(y=-x\). The derivative is positive for \(x<0\) and negative for \(x>0\).

- At the vertex of a smooth parabola, the tangent line is horizontal. - The derivative is positive where the function is increasing.

1. The graph of \(f\) has its minimum at \(x=2\). The tangent line is horizontal there, so \(f^{\prime}(2)=0\). Thus, the derivative graph crosses the x-axis at \(x=2\). 2. For \(x>2\), the graph of \(f\) is increasing. Therefore, its tangent slopes are positive and \(f^{\prime}(x)>0\).

1) \(x=2\) 2) \(f^{\prime}(x)>0\) for \(x>2\).

- Locate the local maximum and local minimum. - The derivative is zero where the tangent line is horizontal. - The derivative is negative where the function is decreasing.

1. The derivative is zero at the local extrema of \(f\). The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\). Therefore, the derivative graph crosses the x-axis at \(x=-2\) and \(x=2\). 2. The graph of \(f\) is decreasing between the local maximum and local minimum. Therefore, \(f^{\prime}(x)<0\) for \(-2<x<2\).

1) \(x=-2\) and \(x=2\) 2) \(f^{\prime}(x)<0\) on \((-2,2)\).

- Find the horizontal tangent line on the parabola. - Estimate tangent slopes at one or two additional inputs. - The derivative of a quadratic function is linear. - Use the increasing and decreasing intervals to check the sign of the derivative graph.

1. The graph of \(f\) has a minimum at \(x=2\), so \(f^{\prime}(2)=0\). 2. From the graph, the tangent slope is about \(-2\) at \(x=0\) and about \(2\) at \(x=4\). 3. Since \(f\) is quadratic, \(f^{\prime}\) is linear. The line through \((0,-2)\), \((2,0)\), and \((4,2)\) is \(f^{\prime}(x)=x-2\).

The derivative graph is the line \(f^{\prime}(x)=x-2\). It crosses the x-axis at \(x=2\) and passes through \((0,-2)\).

- Use the sign of \(f'\) to determine monotonicity. - A local extremum of \(f\) requires a sign change in \(f'\). - Local extrema of \(f'\) correspond to inflection points of \(f\). - The slope of \(f\) at a point is the value of \(f'\) there.

1. Statement a is true. For \(-2<x\le5\), \(f'(x)\le0\), and the derivative is zero only at \(x=3\). Therefore, \(f\) is strictly decreasing there. 2. Statement b is false. At \(x=3\), \(f'\) does not change sign, so \(f\) has no local extremum. 3. Statement c is true. The graph of \(f'\) has a local maximum at \(x=3\), so the concavity of \(f\) changes there. 4. Statement d is false. The graph shows \(f'(1)<0\), so the slope of \(f\) at \(x=1\) is negative.

a) True b) False c) True d) False

- Count the zeros of \(f'\) where the sign changes. - A change from negative to positive gives a local minimum. - Count the local extrema of \(f'\) to determine the number of inflection points of \(f\).

1. Local extrema of \(f\) occur where \(f'\) is zero and changes sign. This happens at \(x = -2\), \(x = 0\), and \(x = 2\), so \(f\) has three local extrema. 2. At \(x = -2\), \(f'\) changes from positive to negative, giving a local maximum. At \(x = 0\), it changes from negative to positive, giving a local minimum. At \(x = 2\), it changes from positive to negative, giving a local maximum. 3. Inflection points of \(f\) occur where \(f'\) has local extrema. The derivative graph has exactly two local extrema in the displayed interval, so \(f\) has two inflection points.

1. Three local extrema 2. A local minimum at \(x = 0\) 3. Two inflection points

- Where does the parabola have a horizontal tangent? - The derivative graph should have a zero at that input. - Find the point and slope of the tangent at the requested input.

1. Graph \(p\) is the parabola \(f\), and graph \(q\) is its derivative \(f'\). 2. Graph \(p\) has a minimum at \(x=0\), where graph \(q\) crosses the x-axis, as expected for a derivative. 3. From the graphs, \(f(2)=0\) and \(f'(2)=1\). 4. The tangent through \((2,0)\) with slope \(1\) is \(y=x-2\).

a) \(p=f\), \(q=f'\); the derivative is zero where the parabola has its minimum. b) \(y=x-2\)

- Compare the extrema of one graph with the zeros of the other. - The derivative is zero at a smooth local maximum or minimum. - Read both the function value and derivative value at \(x=0\).

1. Graph \(p\) is \(f\) and graph \(q\) is \(f'\). 2. At the maxima and minima of \(p\), graph \(q\) crosses the x-axis, consistent with zero derivative at smooth turning points. 3. At \(x=0\), graph \(p\) gives \(f(0)=0\), while graph \(q\) gives \(f'(0)=2\). 4. The tangent through \((0,0)\) with slope \(2\) is \(y=2x\).

a) \(p=f\), \(q=f'\); the zeros of \(q\) occur at extrema of \(p\). b) \(y=2x\)

- Compare the local extremum of one graph with the zero of the other. - Compare increasing and decreasing intervals with the sign of the possible derivative.

1. Graph \(A\) has a local maximum at \(x=0\), and graph \(B\) is zero at \(x=0\). This matches the fact that the derivative is zero at a smooth local extremum. 2. For \(x>0\), graph \(A\) decreases, and graph \(B\) is below the x-axis. This matches the fact that a derivative is negative where its function decreases. 3. Therefore, graph \(A\) represents \(f\), and graph \(B\) represents \(f'\).

Graph A represents \(f\), and graph B represents \(f'\). The zero of graph B matches the local maximum of graph A, and the sign of graph B matches where graph A increases or decreases.

- Decide whether the curve rises or falls as you move right. - Compare the steepness at different points on the graph. - Describe only the interval that is shown; do not assume the graph continues in the same way forever.

1. The graph rises throughout the displayed interval, so the visible tangent slopes are positive. Thus \(f'>0\) and \(f'\) has no zeros in this interval. 2. The graph becomes steeper from left to right, so the tangent slopes increase. Hence \(f'\) is increasing over the displayed interval. 3. Toward the left edge, the curve becomes less steep, with small positive tangent slopes. A finite graph does not establish the behavior as \(x\to-\infty\).

Over the displayed interval, \(f'\) is positive, increasing, and has no zeros. Toward the left edge, its values are small and positive.

- Determine whether each branch is increasing or decreasing. - A decreasing function has negative tangent slopes. - Remember that \(x=0\) is not in the domain.

1. For \(x<0\), the left branch of \(f\) is strictly decreasing. Therefore, every tangent slope is negative and \(f^{\prime}(x)<0\). 2. For \(x>0\), the right branch is also strictly decreasing. Therefore, every tangent slope is negative and \(f^{\prime}(x)<0\). 3. Thus, the derivative graph lies below the x-axis at every input in the domain, \(x\ne0\).

For both \(x<0\) and \(x>0\), \(f^{\prime}(x)<0\).

- Look at whether the right-hand branch rises or falls. - Compare the tangent slopes at small and large positive x-values. - Explain only what the displayed graph supports.

1. On the right-hand branch, the graph rises, so its tangent slopes are positive: \(f'>0\). 2. The graph becomes flatter from left to right, so the positive slopes decrease. Thus \(f'\) decreases while staying above the x-axis. 3. Toward the right edge, the slopes are positive and small. The finite graph alone does not establish a limit as \(x\to\infty\).

For the displayed positive x-values, \(f'>0\) and decreases as \(x\) increases. Toward the right edge, the tangent slopes are small and positive.

- Match horizontal tangents with zeros of the derivative. - Compare where each function increases or decreases with the sign of its derivative. - A stationary inflection point gives a derivative that touches the \(x\)-axis without changing sign.

1. Graph A is an upward-opening parabola with vertex at the origin. Its derivative is an increasing line through the origin, so A matches 3. 2. Graph B is a downward-opening parabola with vertex \((0, 3)\). Its derivative is a decreasing line through the origin, so B matches 2. 3. Graph C is cubic with a stationary inflection point at the origin. Its derivative is nonnegative and equals \(0\) at the origin, so it is an upward-opening parabola with vertex at the origin. Thus, C matches 1.

A \(\rightarrow\) 3; B \(\rightarrow\) 2; C \(\rightarrow\) 1.

- Match local extrema with zeros of the derivative. - Compare where each function increases or decreases with the sign of the derivative. - Recall the derivative of \(\sin x\).

1. Graph A is a parabola with a minimum at \(x=2\). Its derivative must change from negative to positive at \(x=2\), so A matches graph 1. 2. Graph B is \(\sin x\). Its derivative is \(\cos x\), whose graph is 2. 3. Graph C has a local minimum at \(x=-1\) and a local maximum at \(x=1\). It increases between those values and decreases outside them. Its derivative must be positive on \((-1,1)\), negative outside, and zero at \(x=\pm1\). Thus, C matches graph 3.

A \(\rightarrow\) 1; B \(\rightarrow\) 2; C \(\rightarrow\) 3.

- Local extrema of \(g'\) correspond to inflection points of \(g\). - A zero of \(g'\) without a sign change does not produce a local extremum of \(g\). - Does \(g'\) change sign at \(x=2\)? - What combines a change in concavity with a horizontal tangent?

1. Inflection points of \(g\) occur where the graph of \(g'\) has local extrema. 2. The graph of \(g'\) has a local maximum at \(x=0\) and a local minimum at \(x=2\). Therefore, \(g\) has inflection points at \(x=0\) and \(x=2\). 3. At \(x=2\), \(g'(2)=0\), but \(g'\) only touches the x-axis and does not change sign. Thus \(g\) has no local extremum there. Because \(g\) changes concavity and has a horizontal tangent, it has a stationary inflection point at \(x=2\).

a) \(x=0\) and \(x=2\) b) A stationary inflection point occurs at \(x=2\).

- Local extrema of \(f'\) correspond to inflection points of \(f\). - Increasing \(f'\) means \(f\) is concave up. - Decreasing \(f'\) means \(f\) is concave down.

1. The graph of \(f'\) has local extrema at \(x=-2\) and \(x=2\). Therefore, \(f\) has inflection points at those x-values. 2. The displayed x-range is \([-4, 5]\). The graph of \(f'\) is increasing on \([-4, -2)\) and \((2, 5]\), so \(f\) is concave up there. 3. The graph of \(f'\) is decreasing on \((-2, 2)\), so \(f\) is concave down there.

Inflection-point x-coordinates: \(x=-2\) and \(x=2\). Concave up on \([-4,-2)\) and \((2,5]\); concave down on \((-2,2)\).

- Determine where the original function is increasing and decreasing. - The derivative is zero where the original graph has a horizontal tangent line. - Check the sign of each candidate on both sides of the zero.

1. The graph of \(f\) is an upward-opening parabola with a local minimum at \(x=0\). 2. Therefore, \(f^{\prime}(0)=0\). Both candidate graphs have a zero at \(x=0\). 3. For \(x<0\), the graph of \(f\) is decreasing, so \(f^{\prime}(x)<0\). For \(x>0\), it is increasing, so \(f^{\prime}(x)>0\). 4. Graph 1 has this sign pattern, while graph 2 has the opposite sign pattern. Therefore, graph 1 represents \(f^{\prime}\).

Graph 1 represents \(f^{\prime}\).

- What must a derivative equal at a smooth local extremum? - What type of function is the derivative of a line? - Compare the zeros of one graph with the horizontal tangents of the other.

1. The graph of \(f\) has a local minimum at \(x = 2\). If \(g = f'\), then \(g(2)\) would have to be \(0\). Instead, the graph shows \(g(2) > 0\), so \(g\) is not the derivative of \(f\). 2. The graph of \(g\) is a line with constant slope \(-0.5\). Therefore, its derivative is the constant function \(-0.5\). Since \(f\) is a parabola, \(f\) cannot be the derivative of \(g\).

\(g\) is not the derivative of \(f\) because \(f\) has a local minimum at \(x = 2\), but \(g(2) > 0\). \(f\) is not the derivative of \(g\) because the derivative of the line \(g\) must be a constant function, not a parabola.

- The tangent line at the highest point of a smooth parabola is horizontal. - Use symmetry to compare slopes at opposite inputs. - The derivative of a quadratic function is linear.

1. At \(x=0\), the graph has its maximum and the tangent line is horizontal. Therefore, \(f^{\prime}(0)=0\). 2. Because \(f\) is a downward-opening parabola, its tangent slopes decrease steadily as \(x\) increases. By symmetry, the slopes are positive to the left of the vertex and negative to the right. 3. Therefore, the graph of \(f^{\prime}\) is a decreasing line through \((0,0)\). Its vertical scale can be estimated from the slopes of \(f\). 4. The x-coordinate of the vertex of \(f\) is the zero of \(f^{\prime}\).

The slope at \(x=0\) is \(0\). The derivative graph is a decreasing line through \((0,0)\). The vertex of \(f\) occurs at the same input as the zero of \(f^{\prime}\).

- Check for a vertical line of symmetry. - A local maximum or minimum has a horizontal tangent. - Determine whether the graph rises or falls on the stated interval. - Count all x-intercepts, including points where the graph only touches the axis.

1. The graph is symmetric about the y-axis, so (1) is true. 2. The graph has a local minimum at \(x=0\), so \(f'(0)=0\). Thus, (2) is true. 3. On \(2<x\le3.5\), the graph is decreasing, so its slope is negative. Thus, (3) is false. 4. The graph has four x-intercepts on \([-3, 3]\), so (4) is true.

(1) True (2) True (3) False (4) True

- What does the sign of the second derivative tell you about concavity? - Look for zeros of \(f''\) where the graph crosses the x-axis. - Concave-down intervals occur where \(f''\) is negative.

1. Inflection points occur where \(f''\) is zero and changes sign. The graph crosses the x-axis at \(x=-4\), \(x=0\), and \(x=4\), so these are the x-coordinates of the inflection points. 2. On the displayed interval \([-6,6]\), \(f''(x)<0\) on \([-6,-4)\) and \((0,4)\). Therefore, \(f\) is concave down on those intervals.

Inflection-point x-coordinates: \(x=-4\), \(x=0\), and \(x=4\). Concave down on \([-6,-4)\) and \((0,4)\).

- Use the point’s position relative to the x-axis for the sign of \(f(x)\). - Use the tangent slope for the sign of \(f'(x)\). - Look for a change in concavity at \(B\).

1. At \(A\), the graph is on the x-axis, increasing, and concave down. Therefore, \(f(x)=0\), \(f'(x)>0\), and \(f''(x)<0\). 2. At \(B\), the graph is above the x-axis and has a horizontal tangent. It also changes concavity there. Therefore, \(f(x)>0\), \(f'(x)=0\), and \(f''(x)=0\). A point that is both an inflection point and has a horizontal tangent is a stationary inflection point. 3. At \(C\), the graph is above the x-axis, increasing, and concave up. Therefore, \(f(x)>0\), \(f'(x)>0\), and \(f''(x)>0\).

Point \(A\): \(f(x)=0\), \(f'(x)>0\), \(f''(x)<0\) Point \(B\): \(f(x)>0\), \(f'(x)=0\), \(f''(x)=0\); \(B\) is a stationary inflection point. Point \(C\): \(f(x)>0\), \(f'(x)>0\), \(f''(x)>0\)

- Confirm that the displayed graph represents \(f''\), not \(f\) or \(f'\). - Use the sign of \(f''\) to determine concavity. - An inflection point requires a sign change in \(f''\). - Use \(f''\) as the derivative of \(f'\) to analyze the monotonicity of \(f'\).

1. Statement a is true. On \((1,3)\), \(f''(x)>0\), so \(f\) is concave up. 2. Statement b is false. The graph shows \(f''(2)=1\ne0\), so \(x=2\) is not the x-coordinate of an inflection point. 3. Statement c is false. The derivative of \(f'\) is \(f''\). On \([0,2]\), \(f''\) is negative for \(0\le x<1\) and positive for \(1<x\le2\), so \(f'\) is not increasing throughout the interval. 4. Statement d is true. At \(x=3\), \(f''\) changes from positive to negative, so \(f\) changes concavity.

a) True b) False c) False d) True

- Look for points with horizontal tangents. - Inspect the graph's concavity near \(x=0\).

1. Horizontal tangents occur at the visible local extrema. From the graph, they occur at approximately \(x=-2.2\), \(x=0\), and \(x=2.2\). 2. Near \(x=0\), the graph is concave down. Therefore, \(f''(0)<0\).

a) \(x\approx-2.2\), \(x=0\), and \(x\approx2.2\) b) \(f''(0)<0\)

- A smooth local extremum has a horizontal tangent. - Inspect the graph's concavity at \(P\) to determine the sign of the second derivative.

1. The marked point \(P\) at \(x=-2\) is a local maximum, so the tangent is horizontal. Therefore, \(f'(-2)=0\). 2. The graph is concave down at \(x=-2\), so \(f''(-2)<0\).

\(f'(-2)=0\) and \(f''(-2)<0\)

- Use the sign of \(f''\) to determine concavity. - Use \(f''\) as the derivative of \(f'\) to determine whether \(f'\) increases or decreases. - For a polynomial, what must happen to \(f''\) at an inflection point? - Apply the second derivative test in part d.

1. Statement a is true because \(f''(x)<0\) for \(-1<x<3\). 2. Statement b is false because \(f''(x)>0\) for \(3<x\le5\), so \(f'\) is increasing there. 3. Statement c is false because \(f''(1)\ne0\). 4. Statement d is true. If \(f'(0)=0\) and \(f''(0)<0\), the second derivative test gives a local maximum at \(x=0\).

a) True b) False c) False d) True

- What must the second derivative do at an inflection point? - Is a zero of the second derivative alone enough to guarantee an inflection point? - Count the zeros with sign changes in each graph. - Distinguish between crossing and merely touching the x-axis.

1. Inflection points of \(g\) occur where \(g''\) changes sign. 2. In graph III, the curve touches the x-axis at \(x=-3\) and \(x=3\), but it remains negative on both sides of each zero. There are no sign changes, so this graph would produce no inflection points. 3. Graph II crosses the x-axis at \(x=-3\), \(x=0\), and \(x=3\). Each crossing is a sign change, so this graph would produce three inflection points. 4. Graph I crosses the x-axis only at \(x=-3\) and \(x=3\), and it changes sign at both zeros. Therefore, graph I produces exactly two inflection points.

Graph I. It has exactly two zeros with sign changes, at \(x=-3\) and \(x=3\). Graph II has three such zeros, while graph III has no sign changes at its zeros.

- Local maxima and minima of a function correspond to zeros of its derivative. - Compare where each function increases or decreases with where each derivative is positive or negative. - Check all zeros of each possible derivative graph. - Verify that the derivative sign agrees with the slope of the function graph.

1. The graph of \(f\) has two symmetric local minima and a local maximum at \(x=0\). Its derivative must be zero at those three \(x\)-coordinates. Graph \(h_2\) has zeros at the same locations, with signs that match the increasing and decreasing behavior of \(f\). Therefore, \(f\) matches \(h_2\). 2. The graph of \(g\) has two symmetric local maxima and a local minimum at \(x=0\). Graph \(h_1\) has zeros at the same three \(x\)-coordinates, with signs that match the increasing and decreasing behavior of \(g\). Therefore, \(g\) matches \(h_1\).

\(f\) matches \(h_2\), and \(g\) matches \(h_1\).

- The sign of \(f'\) determines increasing and decreasing behavior. - A sign change of \(f'\) classifies a local extremum. - Whether \(f'\) is increasing or decreasing determines concavity. - The y-value of the derivative graph is the tangent slope of \(f\).

1. True. The graph shows \(f'>0\) on \((1,4)\), so \(f\) is strictly increasing there. 2. True. At \(x=5\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 3. False. The derivative graph is increasing on \((0,2)\), so \(f\) is concave up there. 4. True. The tangent slope is \(f'(3)=2\).

(1) True. (2) True. (3) False. (4) True.

- Zeros of a derivative correspond to horizontal tangents of the original function. - A positive derivative means the original function is increasing; a negative derivative means it is decreasing. - Compare both relationships at the quarter-period points.

1. Graph (1) has zeros at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). Graph (2) has a local maximum and a local minimum at those same x-values, as an original function should when its derivative is zero and changes sign. 2. Graph (1) is positive on \((0,\frac{\pi}{2})\), and graph (2) is increasing there. Graph (1) is negative on \((\frac{\pi}{2},\frac{3\pi}{2})\), and graph (2) is decreasing there. 3. Therefore, graph (1) is \(f'\), and graph (2) is \(f\).

Graph (1) is \(f'\); graph (2) is \(f\).

- Compare the y-value of one graph with the slope of the other at the same x-value. - Check whether the sign of the candidate derivative matches the direction of change of the candidate original function. - Use more than one point or interval to confirm the match.

1. If graph (2) is \(f'\), its value at each x-value must equal the slope of graph (1). 2. At \(x=0\), graph (2) has value \(0.5\). Graph (1) has a positive slope of \(0.5\) there. 3. Across the displayed interval, graph (2) stays positive while graph (1) is increasing, and the changing values of graph (2) agree with the changing steepness of graph (1). Therefore, graph (1) is \(f\), and graph (2) is \(f'\).

Graph (1) is \(f\); graph (2) is \(f'\).

- Compare the sign of one graph with whether the other graph rises or falls. - Compare whether the candidate derivative is increasing or decreasing with the concavity of the other graph. - Check the steepness of graph (2) where graph (1) is large and where it is small.

1. Graph (1) stays positive throughout the displayed interval, while graph (2) increases. Thus graph (1) could be the derivative of graph (2). 2. Graph (1) decreases as \(x\) increases. Graph (2) is concave down, as expected when its derivative decreases. 3. Therefore, graph (2) is \(f\), and graph (1) is \(f'\).

Graph (2) is \(f\), and graph (1) is \(f'\).

- Use where the derivative graph lies above or below the x-axis. - At each zero of the derivative, compare its signs on the two sides. - For concavity, focus on whether the derivative graph is decreasing or increasing.

1. Here \(f'(x)=x^2-4\). It is positive on \((-4,-2)\) and \((2,4)\), and negative on \((-2,2)\). Thus \(f\) increases on \((-4,-2)\) and \((2,4)\), and decreases on \((-2,2)\). 2. At \(x=-2\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. At \(x=2\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 3. Since \(f''(x)=2x\), \(f''<0\) for \(x<0\) and \(f''>0\) for \(x>0\). Therefore, \(f\) is concave down on \((-4,0)\), concave up on \((0,4)\), and has an inflection point at \(x=0\).

a) Increasing on \((-4,-2)\) and \((2,4)\); decreasing on \((-2,2)\) b) Local maximum at \(x=-2\); local minimum at \(x=2\) c) Concave down on \((-4,0)\); concave up on \((0,4)\); inflection point at \(x=0\)

- Compare the direction of the curve near the two marked x-values. - Use the function formula to calculate the two tangent slopes. - A graph approaching a horizontal line is not, by itself, proof of a derivative limit.

1. The graph is visibly steeper near \(x=0\) than near \(x=3\). 2. Differentiating gives \(f'(x)=e^{-x}\), so \(f'(0)=1\) and \(f'(3)=e^{-3}<1\), confirming the graph-based comparison. 3. As \(x\to\infty\), \(e^{-x}\to0\), so the derivative approaches \(0\) from above.

Steeper at \(x=0\): \(f'(0)=1\) and \(f'(3)=e^{-3}\). Also, \(f'(x)=e^{-x}\to0\) as \(x\to\infty\).

- Decide which curve could represent tangent slopes of the other curve. - An increasing function has positive derivative values; if it becomes less steep, those derivative values should decrease. - After identifying \(f'\), read its y-values at the requested inputs.

1. Curve \(p\) is \(f\), and curve \(q\) is \(f'\). Curve \(p\) is increasing but becomes less steep, so its derivative should stay positive while decreasing; curve \(q\) has exactly that behavior. 2. From curve \(q\), \(f'(1)=0.5\) and \(f'(4)=0.25\). 3. The function and derivative have the same value where curves \(p\) and \(q\) intersect, which occurs at \(x=0.5\).

a) \(p=f\) and \(q=f'\) b) At \(x=1\), the slope is \(0.5\); at \(x=4\), the slope is \(0.25\). c) \(x=0.5\)

- Identify which graph approaches the x-axis for large positive \(x\). - Compare behavior as \(x\to-\infty\). - Use zeros and y-intercepts to distinguish the functions. - Differentiate when an extremum is a useful identifying feature.

1. For \(g_1(x)=4xe^{-x}\), the graph has a zero at \(x=0\), approaches \(0\) as \(x\to\infty\), and approaches \(-\infty\) as \(x\to-\infty\). Also, \(g_1'(x)=4(1-x)e^{-x}\), so it has a maximum at \(x=1\), with value \(4/e\approx1.47\). This matches Graph A. 2. For \(g_2(x)=e^x-x-1\), \(g_2(0)=0\) and \(g_2'(x)=e^x-1\), which changes from negative to positive at \(x=0\). Thus, \((0, 0)\) is a minimum. The function approaches \(\infty\) in both directions. This matches Graph B. 3. For \(g_3(x)=2-e^{0.5x}\), the graph approaches the horizontal asymptote \(y=2\) as \(x\to-\infty\), has y-intercept \(1\), and approaches \(-\infty\) as \(x\to\infty\). This matches Graph C.

Graph A: \(g_1(x)=4xe^{-x}\); Graph B: \(g_2(x)=e^x-x-1\); Graph C: \(g_3(x)=2-e^{0.5x}\)

- Horizontal tangents reveal zeros of the first derivative. - Read the sign of \(f'\) from whether the graph rises or falls. - Read the sign of \(f''\) from the graph's concavity. - A change in concavity identifies an inflection point.

1. Zeros of \(f'\) occur where the graph of \(f\) has horizontal tangents. The graph has horizontal tangents at \(x=-2\) and \(x=2\). 2. The function decreases between those turning points, so \(f'(x)<0\) on \((-2,2)\). 3. The graph is concave up for \(x>0\), so in the displayed domain \(f''(x)>0\) on \((0,5)\). 4. At \(x=-2\), the graph changes from increasing to decreasing, so \(f\) has a local maximum. At \(x=2\), it changes from decreasing to increasing, so \(f\) has a local minimum. 5. The concavity changes at \(x=0\), so \(f\) has an inflection point there.

a) \(x=-2\) and \(x=2\) b) \((-2,2)\) c) \((0,5)\) d) Local maximum at \(x=-2\); local minimum at \(x=2\) e) Inflection point at \(x=0\)

- Mark where the given derivative is positive, zero, and negative. - Translate those signs into increasing and decreasing behavior of the original function. - Use the derivative's zeros to locate the original function's horizontal tangents.

1. The derivative is positive on \((0,\frac{\pi}{2})\), zero at \(x=\frac{\pi}{2}\), negative on \((\frac{\pi}{2},\frac{3\pi}{2})\), zero at \(x=\frac{3\pi}{2}\), and positive on \((\frac{3\pi}{2},2\pi)\). 2. Therefore, \(f\) must increase, then have a local maximum at \(x=\frac{\pi}{2}\), decrease to a local minimum at \(x=\frac{3\pi}{2}\), and then increase again. 3. Graph a has exactly this behavior. Thus graph a could represent \(f\).

Graph a

- Find where the given derivative is zero. - Use the sign of the derivative on each side of that zero. - Use whether the derivative itself is increasing or decreasing to determine concavity.

1. The derivative has a zero at \(x=2\). 2. For \(x<2\), \(f'(x)<0\), so \(f\) decreases. For \(x>2\), \(f'(x)>0\), so \(f\) increases. Therefore, \(f\) has a local minimum at \(x=2\). 3. Since \(f'\) is linear with positive slope, \(f\) must be concave up. Only graph a is concave up and has its minimum at \(x=2\).

Graph a

- A zero of the derivative must be interpreted using the derivative's signs on both sides. - Keep the derivative's sign separate from whether the derivative itself is rising or falling. - Use one feature of the graph for extrema and a different feature for concavity.

1. Claim 1 is incorrect. At \(x=-2\), \(q'\) changes from positive to negative, so \(q\) has a local maximum, not a local minimum. 2. Claim 2 is incorrect. Concavity is determined by whether \(q'\) is increasing or decreasing, not by the sign of \(q'\). The derivative decreases on \((-5,0)\), so \(q\) is concave down there; it increases on \((0,5)\), so \(q\) is concave up there. Thus \(x=0\) is the x-coordinate of an inflection point.

1. Incorrect: \(q\) has a local maximum at \(x=-2\). 2. Incorrect: \(q\) is concave down on \((-5,0)\) and concave up on \((0,5)\); \(x=0\) is an inflection-point x-coordinate.

- Combine the sign of \(f'\) and the sign of \(f''\) on each interval. - Split the blueprint at \(-2\), \(0\), and \(1\). - Use the supplied function values only to place the named landmarks.

1. On \(( -\infty,-2)\), \(f'>0\) and \(f''<0\): increasing, concave down. 2. At \(x=-2\), \(f'\) changes positive to negative, so \((-2,4)\) is a local maximum. 3. On \((-2,0)\), \(f'<0\) and \(f''<0\): decreasing, concave down. 4. At \(x=0\), concavity changes from down to up, so \((0,1)\) is an inflection point. 5. On \((0,1)\), \(f'<0\) and \(f''>0\): decreasing, concave up. 6. At \(x=1\), \(f'\) changes negative to positive, so \((1,0)\) is a local minimum. 7. On \((1,\infty)\), \(f'>0\) and \(f''>0\): increasing, concave up.

Increasing/concave down on \(( -\infty,-2)\); local max \((-2,4)\); decreasing/concave down on \((-2,0)\); inflection \((0,1)\); decreasing/concave up on \((0,1)\); local min \((1,0)\); increasing/concave up on \((1,\infty)\).

- Use the zeros and signs of the first derivative for the rise-fall pattern. - Use the second derivative graph separately to locate changes in bending. - Combine both descriptions into a single graph shape; the actual vertical position is unknown.

1. The first graph represents \(f'(x)=x^2-4\), with zeros \(-2\) and \(2\). It is positive outside these roots and negative between them. 2. Thus \(f\) increases on \((-3,-2)\), decreases on \((-2,2)\), and increases on \((2,3)\). It has a local maximum at \(-2\) and a local minimum at \(2\). 3. The second graph represents \(f''(x)=2x\), negative to the left of \(0\) and positive to the right. 4. Hence \(f\) is concave down on \((-3,0)\), concave up on \((0,3)\), and changes concavity at \(x=0\).

Increasing: \((-3,-2)\) and \((2,3)\). Decreasing: \((-2,2)\). Local maximum at \(x=-2\), local minimum at \(x=2\). Concave down for \(x<0\), concave up for \(x>0\), with an inflection point at \(x=0\).

- The second derivative describes how the first derivative changes. - Look at the direction of the first derivative graph on the right of the origin. - Compare that direction with the sign shown by the second derivative graph.

1. The first graph shows \(f'(x)=x^2+1\), which decreases on \((-2,0)\) and increases on \((0,2)\). 2. The second graph claims \(f''(x)=-2\) everywhere, which would make \(f'\) strictly decreasing throughout the interval. 3. These two properties conflict on \((0,2)\), so the graphs cannot be derivatives of the same function.

No. The graph of \(f'\) increases on \((0,2)\), but a negative \(f''\) would require \(f'\) to decrease there.

- Does the sign of the first derivative change at the origin? - What happens to the sign of the second derivative there? - Combine movement and bending before naming the point.

1. The graph of \(f'\) is \(3x^2\). It is positive on each side of \(0\) and zero at \(0\). 2. Thus \(f\) increases on both sides of \(0\) and has a horizontal tangent there, not a local maximum or minimum. 3. The second derivative graph is \(6x\), negative to the left of \(0\) and positive to the right. 4. The graph of \(f\) changes from concave down to concave up at \(0\), so it has a stationary inflection point.

\(f\) rises through a horizontal tangent at \(x=0\), changing from concave down to concave up. It has a stationary inflection point, not a local extremum.

- First identify the direction of the graph on each stated interval. - Compare values at the ends of a single monotone interval. - Could a rise, a fall, and another rise combine to give different overall changes?

1. On \((-3,-1)\), positive derivative means \(f\) increases; continuity at the ends gives \(f(-1)>f(-3)\). 2. On \((-1,1)\), negative derivative means \(f\) decreases, so \(f(-1)>f(1)\). 3. On \((1,3)\), positive derivative means \(f(3)>f(1)\). 4. The signs alone do not compare \(f(-3)\) and \(f(3)\). The total rise and fall can have different sizes, so part d) need not hold.

a) Must be true. b) Must be true. c) Must be true. d) Not necessarily true.

- How could zeros and changes in direction help locate the features of the sketch? - What information does each derivative contribute to the graph description? - Which parts of the domain need special attention when naming graph features?

1. Factor \(f(x)=(x-3)(x+1)\), giving x-intercepts at \(x=-1\) and \(x=3\). 2. Differentiate: \(f'(x)=2x-2\), which vanishes at \(x=1\). Its sign is negative before \(1\) and positive after, so the vertex is the minimum \((1,-4)\). 3. The graph decreases on \((-1,1)\) and increases on \((1,4)\). Since \(f''(x)=2>0\), it is concave up throughout. 4. The endpoint values are \(f(-1)=0\) and \(f(4)=5\).

a) Vertex \((1,-4)\); x-intercepts \((-1,0)\) and \((3,0)\). b) \(f'(x)=2x-2\) is negative on \((-1,1)\) and positive on \((1,4)\), so the curve decreases and then increases. Since \(f''(x)=2>0\), it is concave up throughout. c) Left endpoint \((-1,0)\); right endpoint \((4,5)\).

- Read the two derivative signs in each row as two different graph features. - A turning point depends on a change in the first derivative's sign. - An inflection point depends on a change in concavity. - Use the supplied points only after deciding which kind of feature each one has.

1. Positive to negative \(f'\) at \(-2\) gives a local maximum at \((-2,4)\). 2. Negative to positive \(f'\) at \(1\) gives a local minimum at \((1,0)\). 3. The sign of \(f''\) changes from negative to positive at \(x=0\), so \((0,2)\) is an inflection point. 4. The first interval is increasing and concave down; the second is decreasing and concave down; the third is decreasing and concave up; the fourth is increasing and concave up.

Local maximum \((-2,4)\); local minimum \((1,0)\); inflection point \((0,2)\). On \((-3,-2)\): increasing, concave down. On \((-2,0)\): decreasing, concave down. On \((0,1)\): decreasing, concave up. On \((1,2)\): increasing, concave up.

- The plotted curve is the derivative, so its height tells you the sign of \(f'\). - Its rise or fall tells you the concavity of \(f\). - Compare the function values by walking along an interval where the function decreases.

1. The derivative is positive to the left of \(-1\), negative between \(-1\) and \(1\), and positive to the right of \(1\). Thus \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=1\). 2. The derivative graph decreases up to \(x=0\) and increases afterward, so concavity changes at \(x=0\). 3. Since \(f\) decreases from \(-1\) to \(0\), \(f(-1)>f(0)=1\). Since it also decreases from \(0\) to \(1\), \(f(1)<1\).

a) Local maximum at \(x=-1\); local minimum at \(x=1\). b) Inflection-point x-coordinate \(0\). c) \(f(-1)>1\) and \(f(1)<1\).

- Think of a familiar cubic that flattens but keeps increasing. - Moving a curve up or down changes where it crosses the y-axis but not its slopes. - Check a horizontal tangent and a concavity change separately.

1. One choice is \(f(x)=x^3+2\), which passes through \((0,2)\). 2. Its derivative is \(f'(x)=3x^2\), nonnegative everywhere and positive except at \(0\), so the function is strictly increasing. 3. Its second derivative is \(f''(x)=6x\), which changes from negative to positive at \(0\), giving an inflection point. 4. Since \(f'(0)=0\), the tangent at that point is horizontal.

One example is \(f(x)=x^3+2\), with \(f'(x)=3x^2\) and \(f''(x)=6x\).

- Treat each formula as a separate graph-analysis problem before matching any properties. - Use the sign of the first derivative for monotonicity and local extrema. - A stationary inflection point needs both a horizontal tangent and a concavity change. - Use algebraic symmetry and factoring only for the symmetry and zero parts of the blueprint.

1. For \(f(x)=2x^2-x^4\), the graph is y-axis symmetric and the zeros are \(-\sqrt2,0,\sqrt2\). Since \(f'(x)=4x(1-x^2)\), \(f\) increases on \(( -\infty,-1)\) and \((0,1)\), and decreases on \((-1,0)\) and \((1,\infty)\). Thus the local maxima are \((-1,1)\) and \((1,1)\), and the local minimum is \((0,0)\). Its inflection points are not stationary. Therefore A and E apply. 2. For \(g(x)=x^3-3x\), the graph is origin-symmetric and the zeros are \(-\sqrt3,0,\sqrt3\). Since \(g'(x)=3(x^2-1)\), \(g\) increases on \(( -\infty,-1)\) and \((1,\infty)\), and decreases on \((-1,1)\). Thus \((-1,2)\) is a local maximum and \((1,-2)\) is a local minimum. Its inflection point at \(x=0\) is not stationary. Therefore B, C, and E apply. 3. Since \(h(x)=(x-1)^3\), its only zero is \(x=1\). The derivative \(h'(x)=3(x-1)^2\) is nonnegative and does not change sign, so \(h\) is strictly increasing on \(\mathbb{R}\) and has no local extremum. Because \(h''(x)=6(x-1)\) changes sign at \(x=1\) and \(h'(1)=0\), \((1,0)\) is a stationary inflection point. Therefore D applies.

(1) Blueprint: y-axis symmetry; zeros \(-\sqrt2,0,\sqrt2\); increasing on \(( -\infty,-1)\), \((0,1)\); decreasing on \((-1,0)\), \((1,\infty)\); local maxima \((-1,1)\), \((1,1)\); local minimum \((0,0)\); no stationary inflection point. Properties A, E. (2) Blueprint: origin symmetry; zeros \(-\sqrt3,0,\sqrt3\); increasing on \(( -\infty,-1)\), \((1,\infty)\); decreasing on \((-1,1)\); local maximum \((-1,2)\); local minimum \((1,-2)\); no stationary inflection point. Properties B, C, E. (3) Blueprint: zero \(1\); strictly increasing on \(\mathbb{R}\); no local extrema; stationary inflection point \((1,0)\). Property D.

- Factor or complete the square before differentiating when that reveals a useful graph feature. - A zero of the first derivative does not automatically create a local extremum; check the sign on both sides. - For a stationary inflection point, verify a concavity change at a horizontal tangent. - Assemble all of the features for one function before matching its statements.

1. For \(f(x)=x^4+1\), \(f'(x)=4x^3\) has the single zero \(x=0\). The function decreases on \(( -\infty,0)\), increases on \((0,\infty)\), and has a local minimum at \((0,1)\). It has no real zeros and no stationary inflection point. Thus only A applies. 2. Since \(g(x)=(x+1)^3\), \(g'(x)=3(x+1)^2\) has the single zero \(x=-1\) but does not change sign. The function is strictly increasing on \(\mathbb{R}\), has no local extremum, and \(g''(x)=6(x+1)\) changes sign at \(-1\). Thus \((-1,0)\) is a stationary inflection point. A, B, and C apply. 3. For \(h(x)=(x-1)^2+4\), \(h'(x)=2(x-1)\) has the single zero \(x=1\). The function decreases on \(( -\infty,1)\), increases on \((1,\infty)\), and has a local minimum at \((1,4)\). It has no real zeros and is symmetric about \(x=1\). Thus A and D apply.

(1) Blueprint: no real zeros; decreasing then increasing at local minimum \((0,1)\); no stationary inflection point. Statement A. (2) Blueprint: zero and stationary inflection point \((-1,0)\); strictly increasing on \(\mathbb{R}\); no local extremum. Statements A, B, C. (3) Blueprint: no real zeros; decreasing on \(( -\infty,1)\), increasing on \((1,\infty)\); local minimum \((1,4)\); symmetry line \(x=1\). Statements A, D.

- Even powers suggest y-axis symmetry. - Factor out the greatest common factor to find the zeros. - Use the first and second derivatives to find and classify extrema. - The leading term determines the graph's end behavior.

1. Because \(f(-x)=f(x)\), the graph is symmetric about the y-axis. 2. Factoring gives \(f(x)=x^2(x-2)(x+2)\). The zeros are \(x=0\), with multiplicity \(2\), and \(x=\pm2\), each with multiplicity \(1\). 3. The derivative is \(f'(x)=4x(x^2-2)\), so the critical numbers are \(x=0\) and \(x=\pm\sqrt{2}\). The second derivative is \(f''(x)=12x^2-8\). Thus, \((0, 0)\) is a local maximum, and \((\pm\sqrt{2}, -4)\) are local minima. 4. The leading term is \(x^4\), so \(f(x)\to\infty\) as \(x\to\pm\infty\). The graph passes through the intercepts and extrema, is symmetric about the y-axis, and has both ends rising.

1. The graph is symmetric about the y-axis. 2. The zeros are \(x=0\), with multiplicity \(2\), and \(x=\pm2\), each with multiplicity \(1\). 3. Local maximum: \((0, 0)\); local minima: \((-\sqrt{2}, -4)\) and \((\sqrt{2}, -4)\) 4. \(f(x)\to\infty\) as \(x\to\pm\infty\).

- Substitute \(u=x^2\) to turn the quartic equation into a quadratic. - Remember both square roots when substituting back. - Use the first and second derivatives to find the extrema. - Use symmetry and end behavior to complete the graph description.

1. Set \(f(x)=0\) and multiply by \(-2\): \(x^4-8x^2+12=0\). Let \(u=x^2\). Then \(u^2-8u+12=(u-2)(u-6)=0\), so \(u=2\) or \(u=6\). Therefore, the x-intercepts occur at \(x=\pm\sqrt{2}\) and \(x=\pm\sqrt{6}\). 2. The derivative is \(f'(x)=-2x(x^2-4)\), so the critical numbers are \(x=-2\), \(x=0\), and \(x=2\). Since \(f''(x)=-6x^2+8\), \((0, -6)\) is a local minimum and \((\pm2, 2)\) are local maxima. 3. The graph is symmetric about the y-axis. It passes through the four x-intercepts, the local minimum, and the two local maxima. Since the leading coefficient is negative, both ends fall toward \(-\infty\).

1. The x-intercepts occur at \(x=\pm\sqrt{2}\) and \(x=\pm\sqrt{6}\). 2. Local minimum: \((0, -6)\); local maxima: \((-2, 2)\) and \((2, 2)\) 3. The graph is symmetric about the y-axis, passes through the listed intercepts and extrema, and satisfies \(f(x)\to-\infty\) as \(x\to\pm\infty\).

- Would an algebraically equivalent form of \(f\) make differentiation easier? - What information about the curve at \(P\) determines its tangent line? - Can the derivative have the same sign on both sides of a vertical asymptote?

1. Rewrite \(f(x)=\frac{9x^2-12x+4}{2x}=4.5x-6+\frac{2}{x}\), for \(x\ne0\). 2. Differentiate: \(f'(x)=4.5-\frac{2}{x^2}=\frac{9x^2-4}{2x^2}\). 3. At \(x=1\), the slope is \(f'(1)=2.5\). Through \(P\), the tangent is \(y-0.5=2.5(x-1)\), or \(y=2.5x-2\). 4. The derivative has a vertical asymptote at \(x=0\), since \(9x^2-4\) approaches \(-4\) and \(2x^2\) approaches \(0\). 5. Close to \(0\) on either side, \(9x^2-4<0\) and \(2x^2>0\). Therefore, \(f'(x)<0\) on both sides; there is no sign change near the asymptote.

a) \(y=2.5x-2\) b) The vertical asymptote is \(x=0\). The derivative is negative on both sides close to \(0\), so its sign does not change there.

- Rewrite everything in powers of \(2^x\) before solving the zero and critical-number equations. - The signs of the first and second derivatives determine the monotonicity and concavity pieces of the blueprint. - Use the limiting behavior to decide how the far-left and far-right pieces of the graph must continue. - Order the significant x-values only after computing each feature.

1. Rewrite \(4^x=(2^x)^2\). Then \(f(x)=2^x(2^x-1)\), so the only zero is \(x=0\). 2. \(f'(x)=\ln(2)2^x(2\cdot2^x-1)\). Thus the only critical number is \(x=-1\); \(f'\) is negative before \(-1\) and positive after it, so the point \((-1,-\tfrac14)\) is a local minimum. 3. \(f''(x)=(\ln2)^2 2^x(4\cdot2^x-1)\). It changes sign at \(x=-2\), giving the inflection point \((-2,-\tfrac{3}{16})\). The graph is concave down on \(( -\infty,-2)\) and concave up on \((-2,\infty)\). 4. As \(x\to\infty\), \(f(x)\to\infty\). As \(x\to-\infty\), \(f(x)\to0\) from below. 5. Therefore the blueprint is: approach \(y=0\) from below, concave down and decreasing; pass through the inflection point at \(x=-2\); continue decreasing concave up to the local minimum at \(x=-1\); then increase concave up through the zero at \(x=0\) and toward \(\infty\).

a) Zero: \((0,0)\). b) Local minimum: \((-1,-\tfrac14)\). c) Inflection point: \((-2,-\tfrac{3}{16})\); concave down on \(( -\infty,-2)\), concave up on \((-2,\infty)\). d) \(f(x)\to\infty\) as \(x\to\infty\); \(f(x)\to0^-\) as \(x\to-\infty\). e) Decreasing on \(( -\infty,-1)\), increasing on \((-1,\infty)\); left-to-right landmarks: inflection at \(x=-2\), minimum at \(x=-1\), zero at \(x=0\).

- Positivity of both exponential terms settles the zero question immediately. - Compare the two positive exponential terms in the first derivative to locate the turning point. - The sign of the second derivative tells whether the graph ever changes concavity. - A graph blueprint should connect the two ends through the extremum using the derivative information.

1. Both terms are positive for every real \(x\), so \(g(x)>0\) and there are no zeros. 2. \(g'(x)=\ln(2)(2^x-4\cdot2^{-x})\). Solving \(g'(x)=0\) gives \(x=1\). The derivative is negative for \(x<1\) and positive for \(x>1\), so \((1,4)\) is a local minimum. 3. \(g''(x)=(\ln2)^2(2^x+4\cdot2^{-x})>0\) for every real \(x\), so the graph is concave up everywhere. 4. As \(x\to\infty\), the term \(2^x\) makes \(g(x)\to\infty\). As \(x\to-\infty\), the term \(4\cdot2^{-x}\) makes \(g(x)\to\infty\). 5. Thus the graph descends from \(\infty\), remains concave up, reaches its minimum at \((1,4)\), and then rises to \(\infty\), staying above the x-axis throughout.

a) No zeros because \(g(x)>0\) for all \(x\). b) Local minimum: \((1,4)\). c) Concave up on \(\mathbb{R}\). d) \(g(x)\to\infty\) as \(x\to\pm\infty\). e) Decreasing and concave up on \(( -\infty,1)\); minimum \((1,4)\); increasing and concave up on \((1,\infty)\).

- Locate the local maximum and minimum of \(f\). - What is the derivative value at a smooth extremum? - Determine where \(f\) is increasing and decreasing. - Match those intervals to the sign of each candidate graph.

1. The graph of \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=1\). Therefore, \(f'(-1)=0\) and \(f'(1)=0\). This eliminates graph \(C\), which has only one zero. 2. The function increases for \(x<-1\) and \(x>1\), so \(f'(x)>0\) there. It decreases for \(-1<x<1\), so \(f'(x)<0\) there. 3. Only graph \(A\) has zeros at \(-1\) and \(1\), is negative between them, and is positive outside them. Therefore, graph \(A\) represents \(f'\).

Graph \(A\) represents \(f'\).

- The derivative gives the tangent slope at each input. - At a smooth local maximum or minimum, the tangent is horizontal. - Use a slope triangle on the red tangent line. - Identify the interval where the graph moves downward from left to right.

1. The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\). The tangents are horizontal there, so \(f'(-2)=0\) and \(f'(2)=0\). 2. The tangent \(t\) passes through \((0,0)\) and \((1,-3)\). Its slope is \(\frac{-3-0}{1-0}=-3\), so \(f'(0)=-3\). 3. The graph decreases from the local maximum to the local minimum. Thus, the largest interval is \([-2,2]\).

a) \(x=-2\) and \(x=2\) b) \(f'(0)=-3\) c) \([-2,2]\)

- What does the sign of \(f'\) tell you about whether \(f\) is increasing or decreasing? - What sign change of \(f'\) indicates a local maximum or minimum of \(f\)? - How does the slope of the graph of \(f'\) relate to \(f''\)?

1. Statement a is true. On \(-1<x<1\), the graph of \(f'\) is above the x-axis, so \(f'(x)>0\). Therefore, \(f\) is strictly increasing there. 2. Statement b is false. At \(x=3\), \(f'\) changes from negative to positive. Therefore, \(f\) has a local minimum, not a local maximum. 3. Statement c is true. On \(0<x<2\), the graph of \(f'\) is decreasing. Therefore, \(f''(x)<0\), so \(f\) is concave down there.

a) True; \(f'(x)>0\) on \(-1<x<1\). b) False; \(f'\) changes from negative to positive at \(x=3\), so \(f\) has a local minimum. c) True; \(f'\) is decreasing on \(0<x<2\), so \(f''(x)<0\).

- Compare symmetry, intercepts, end behavior, or extrema for part a. - Use the leading term to determine the end behavior. - Substitute \(u=x^2\) to solve the quartic equation. - Use the first and second derivatives to find and classify extrema.

1. The graph shown cannot represent \(f\). For example, \(f\) is even and therefore symmetric about the y-axis, while the displayed graph is symmetric about \(x=1\). Also, \(f(0)=6\), but the displayed graph has y-intercept \((0, 1)\). 2. The positive leading coefficient and even degree give \(f(x)\to\infty\) as \(x\to\pm\infty\). 3. Solving \(\frac{1}{2}x^4-4x^2+6=0\) and substituting \(u=x^2\) gives \(u^2-8u+12=0\). Thus, \(u=2\) or \(u=6\), so the zeros are \(x=\pm\sqrt{2}\approx\pm1.41\) and \(x=\pm\sqrt{6}\approx\pm2.45\). 4. The derivative is \(f'(x)=2x(x^2-4)\), so the critical numbers are \(x=-2, 0, 2\). Since \(f''(x)=6x^2-8\), \((0, 6)\) is a local maximum and \((-2, -2)\) and \((2, -2)\) are local minima.

a) Possible reasons include: \(f\) is symmetric about the y-axis, but the displayed graph is symmetric about \(x=1\); and \(f(0)=6\), but the displayed graph has y-intercept \((0, 1)\). b) 1. \(f(x)\to\infty\) as \(x\to\pm\infty\) 2. \(x=\pm\sqrt{2}\approx\pm1.41\) and \(x=\pm\sqrt{6}\approx\pm2.45\) 3. Local maximum: \((0, 6)\); local minima: \((-2, -2)\) and \((2, -2)\)

- Use zeros and sign changes of the derivative for local extrema. - For concavity, track whether the derivative graph is increasing or decreasing. - A turning point of \(f'\) corresponds to a concavity change of \(f\) when that direction changes.

1. Local extrema of \(f\) occur at zeros of \(f'\) where the sign changes. 2. At \(x=-1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 3. At \(x=3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 4. The derivative graph has a local minimum at \(x=1\), changing from decreasing to increasing. Thus \(f\) changes from concave down to concave up there, so \(x=1\) is an inflection point.

a) Local maximum at \(x=-1\); local minimum at \(x=3\). b) Inflection point at \(x=1\).

- Compare the zeros of one graph with the horizontal tangents of the other. - A function decreases where its derivative is negative. - An inflection point of \(f\) corresponds to a local extremum of \(f'\). - Read the requested coordinates from the graph of \(f\).

1. The graph labeled \(g_1\) has horizontal tangents at \(x=-2\) and \(x=2\), while the graph labeled \(g_2\) is zero at those values. Also, \(g_1\) decreases between \(-2\) and \(2\), where \(g_2\) is negative. Therefore, \(g_1\) represents \(f\) and \(g_2\) represents \(f'\). 2. Reading from \(g_1\), the local maximum is \((-2, 3)\), the local minimum is \((2, -1)\), and the inflection point is \((0, 1)\). 3. Since \(g_2\) is negative for \(-2<x<2\), \(f\) is strictly decreasing on \([-2, 2]\).

a) \(g_1\) represents \(f\), and \(g_2\) represents \(f'\). b) Local maximum: \((-2, 3)\); local minimum: \((2, -1)\); inflection point: \((0, 1)\) c) \([-2, 2]\)

- Use the sign of \(f'\) to determine where \(f\) increases or decreases. - Check the direction of each sign change at a zero of \(f'\). - A local extremum of \(f'\) indicates a possible inflection point of \(f\).

1. The derivative is positive on \((-4, -2)\) and \((2, 4)\), so \(f\) is strictly increasing on \([-4, -2]\) and \([2, 4]\). The derivative is negative on \((-2, 2)\), so \(f\) is strictly decreasing on \([-2, 2]\). 2. At \(x = -2\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. At \(x = 2\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 3. The graph of \(f'\) has a local minimum at \(x = 0\), so \(f''\) changes sign there. Therefore, the concavity of \(f\) changes at \(x = 0\), giving an inflection point.

a) Increasing on \([-4, -2]\) and \([2, 4]\); decreasing on \([-2, 2]\) b) Local maximum at \(x = -2\); local minimum at \(x = 2\) c) \(f'\) has a local minimum at \(x = 0\), so \(f\) changes concavity there.

- Use the sign of \(f'\) for increasing and decreasing intervals. - Use sign changes of \(f'\) to classify local extrema. - Inflection points of \(f\) occur where \(f'\) changes from increasing to decreasing or vice versa. - The graph of \(f\) is concave up where \(f'\) is increasing and concave down where \(f'\) is decreasing.

1. The derivative is positive on \((-5, -4)\) and \((-1, 4)\), so \(f\) is strictly increasing on \([-5, -4]\) and \([-1, 4]\). It is negative on \((-4, -1)\) and \((4, 5)\), so \(f\) is strictly decreasing on \([-4, -1]\) and \([4, 5]\). 2. At \(x = -4\), \(f'\) changes from positive to negative, giving a local maximum. At \(x = -1\), it changes from negative to positive, giving a local minimum. At \(x = 4\), it changes from positive to negative, giving another local maximum. 3. Inflection points of \(f\) occur where \(f'\) has local extrema. From the graph, these occur at about \(x \approx -2.7\) and \(x = 2\). Using these estimates, \(f'\) increases approximately on \((-2.7, 2)\), so \(f\) is concave up there. It decreases approximately on \((-5, -2.7)\) and \((2, 5)\), so \(f\) is concave down there.

a) Increasing on \([-5, -4]\) and \([-1, 4]\); decreasing on \([-4, -1]\) and \([4, 5]\) b) Local maxima at \(x = -4\) and \(x = 4\); local minimum at \(x = -1\) c) Inflection points at approximately \(x \approx -2.7\) and \(x = 2\); concave up approximately on \((-2.7, 2)\); concave down approximately on \((-5, -2.7)\) and \((2, 5)\)

- Match horizontal tangents of each function with zeros of its derivative. - A derivative is positive where its function increases and negative where its function decreases. - Pay attention to the number and locations of local extrema.

1. Graph A is an upward-opening parabola with a minimum at \(x = 1\). It decreases for \(x < 1\) and increases for \(x > 1\), so its derivative must be an increasing line that is zero at \(x = 1\). Therefore, A matches (2). 2. Graph B has a local maximum at \(x = -2\) and a local minimum at \(x = 2\). Its derivative must be zero at \(x = -2\) and \(x = 2\), negative between those values, and positive outside them. Therefore, B matches (3). 3. Graph C has a maximum at \(x = 0\), increases for \(x < 0\), and decreases for \(x > 0\). Its derivative must be positive to the left of \(0\), zero at \(0\), and negative to the right. Therefore, C matches (1).

A \(\rightarrow\) (2); B \(\rightarrow\) (3); C \(\rightarrow\) (1)

- The derivative value equals the tangent slope. - Use two points on \(t\) to calculate its slope. - The derivative is zero at smooth local extrema. - Match the sign of each candidate to the increasing and decreasing intervals of \(f\).

1. The tangent passes through \(P(2,2)\) and \((4,0)\). Its slope is \(\frac{0-2}{4-2}=-1\), so \(f'(2)=-1\). 2. All three candidates have value \(-1\) at \(x=2\), so that value alone does not determine the answer. 3. The graph of \(f\) has a local maximum near \(x=0.85\) and a local minimum near \(x=3.15\). Therefore, \(f'\) must have zeros near those inputs. 4. Only \(g_1\) has zeros near \(0.85\) and \(3.15\). It is also positive where \(f\) increases and negative where \(f\) decreases. Thus, \(g_1\) represents \(f'\).

a) The slope is \(-1\), so \(f'(2)=-1\). b) \(g_1\) represents \(f'\).

- Find the x-coordinates of the local maximum and minimum. - The derivative must be zero at those inputs. - Use the increasing and decreasing intervals to determine the derivative's sign pattern. - Estimate one slope to distinguish candidates with the same zeros.

1. The graph of \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=3\). Therefore, \(f'\) must have zeros at \(-1\) and \(3\). This eliminates option D. 2. The function increases for \(x<-1\) and \(x>3\), so \(f'\) must be positive there. It decreases between the extrema, so \(f'\) must be negative there. Thus, its graph opens upward, eliminating option B. 3. Options A and C have the correct zeros and sign pattern. Estimate the slope of \(f\) near the inflection point \(x=1\). The graph suggests a slope near \(-2\). 4. Option A gives \(f'(1)=0.5-1-1.5=-2\), while option C gives \(f'(1)=1-2-3=-4\). Therefore, option A matches the graph.

A: \(f'(x)=0.5x^2-x-1.5\)

- Use the sign of \(f'\) to determine whether \(f\) increases or decreases. - What does a sign-changing zero of the derivative indicate? - Compare function values using monotonicity. - How are extrema of \(f'\) related to inflection points of \(f\)?

1. At \(x=0\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. Statement a is true. 2. On \((0, 4)\), \(f'(x)>0\), so \(f\) is strictly increasing, not decreasing. Statement b is false. 3. Since \(f'(x)>0\) on \((0, 3]\), \(f(3)>f(0)=2\). Statement c is true. 4. An extremum of \(f'\) corresponds to an inflection point of \(f\). Since \(f'\) has a local maximum at \(x=2\), statement d is true.

a) True b) False c) True d) True

- First relate \(f(x)\), \(f'(x)\), and \(f''(x)\) to the graph. - The function value is the point’s vertical coordinate. - The first derivative gives the tangent slope. - A negative second derivative corresponds to concave-down behavior. - An inflection point occurs where the graph changes concavity.

1. Compare the y-coordinates of all seven marked points. The lowest marked point is at \(x_6\), where \(f(x_6)\approx-1.54\). 2. The first derivative is the tangent slope. The graph is nearly horizontal at \(x_2\), \(x_4\), and \(x_6\), decreasing at \(x_5\), and increasing steeply at \(x_3\). Therefore, the greatest slope among the five interior values occurs at \(x_3\). 3. A negative second derivative means the graph is concave down. The graph is concave down between its inflection-point x-coordinates \(x=-1\) and \(x=2\). Of the marked values, only \(x_4\) lies strictly between them. 4. The graph changes concavity at \(x_3=-1\) and \(x_5=2\). Therefore, the inflection points correspond to \(x_3\) and \(x_5\).

a) \(x_6\) b) \(x_3\) c) \(x_4\) d) \(x_3\) and \(x_5\)

- Review how monotonicity, concavity, extrema, and inflection points of \(f\) appear in the graph of \(f'\). - Connect the sign of \(f'\) to whether \(f\) is increasing or decreasing. - Connect the monotonicity of \(f'\) to the concavity of \(f\). - What sign changes of \(f'\) produce local extrema of \(f\)? - Where does \(f\) have inflection points when the graph of \(f'\) is given?

1. Statement a is true. The graph of \(f'\) is decreasing on \([0, 2]\), so \(f''(x)<0\) there and \(f\) is concave down. 2. Statement b is false. At \(x=1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. The graph of \(f'\) does not have a local extremum there, so \(f\) does not have an inflection point there. 3. Statement c is true. The graph of \(f'\) has local extrema at \(x=1-\sqrt{3}\approx-0.73\) and \(x=1+\sqrt{3}\approx2.73\). Therefore, \(f\) has exactly two inflection points on the displayed interval. 4. Statement d is true. On \((1, 4)\), \(f'(x)<0\), with zeros only at the endpoints. Therefore, \(f\) is strictly decreasing on \([1, 4]\). 5. Statement e is false. At \(x=-2\), \(f'\) changes from negative to positive, so \(f\) has a local minimum.

a) True b) False c) True d) True e) False

- First compute \(f'(x)\). - Compare the degree and leading coefficient with the displayed graph. - Compare signs on intervals where \(f\) increases or decreases. - Compare the directions of sign changes at the zeros.

1. Differentiate: \(f'(x)=x^2-2x-3\). 2. End behavior: \(f'\) is an upward-opening parabola, so \(f'(x)\to\infty\) as \(x\to\pm\infty\). The displayed graph opens downward. 3. Sign and monotonicity: \(f'(x)<0\) on \((-1,3)\), so \(f\) decreases there. The displayed graph is positive on that interval. 4. Critical-number behavior: At \(x=-1\), the true derivative changes from positive to negative, which gives \(f\) a local maximum. The displayed graph changes from negative to positive there, which would indicate a local minimum.

Three valid reasons are: 1. The true derivative \(f'(x)=x^2-2x-3\) opens upward, but the displayed graph opens downward. 2. The true derivative is negative on \((-1,3)\), but the displayed graph is positive there. 3. At \(x=-1\), the true derivative changes from positive to negative, while the displayed graph changes from negative to positive.

- Compare the end behavior of a positive-leading-coefficient cubic with each graph. - Factor \(f\) to identify its zeros and their multiplicities. - Check whether \(f\) is even or odd. - Find the critical numbers and compare the locations and classifications of the extrema.

1. Graph A cannot represent \(f\). The function has end behavior \(f(x)\to\infty\) as \(x\to\infty\) and \(f(x)\to-\infty\) as \(x\to-\infty\), while Graph A has the opposite end behavior. Also, \(f(x)=x^2(x-3)<0\) for every \(x<0\), but Graph A lies above the x-axis there. In addition, \(f\) has a local maximum at \((0, 0)\) and a local minimum at \((2, -4)\), while Graph A reverses those classifications. 2. Graph B cannot represent \(f\). Graph B is symmetric about the origin, but \(f\) is neither even nor odd. Graph B crosses the x-axis at the origin, while \(f(x)=x^2(x-3)\) has a double zero there and only touches the axis. Also, the local maximum of \(f\) is at the origin, not at a negative x-value as shown in Graph B.

Graph A: It has the wrong end behavior; it lies above the x-axis for negative \(x\), although \(f(x)<0\) there; and it reverses the local maximum and local minimum. Graph B: It has origin symmetry, but \(f\) has no such symmetry; it crosses rather than touches the x-axis at \(x=0\); and its local maximum is at the wrong x-value.

- Use zeros and sign changes of \(f'\) to classify critical numbers of \(f\). - At a zero of \(f'\), check whether the derivative changes sign or only touches the x-axis. - How does the monotonicity of \(f'\) determine the concavity of \(f\)? - Inflection-point x-coordinates of \(f\) correspond to local extrema of \(f'\). - Use the sign of \(f'\) to compare \(f(-2)\) and \(f(-4)\).

1. At \(x=-4\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x=-1\), \(f'\) has a double zero and does not change sign, so \(f\) has no local extremum there; instead, it has a stationary inflection point. 2. On the displayed interval, \(f'\) is increasing on \([-5, -3)\) and \((-1, 0.5]\), so \(f\) is concave up there. It is decreasing on \((-3, -1)\), so \(f\) is concave down there. 3. The graph of \(f'\) has local extrema at \(x=-3\) and \(x=-1\). Therefore, these are the x-coordinates of the inflection points of \(f\). 4. For \(-4<x\le-2\), \(f'(x)>0\), so \(f\) is strictly increasing on \([-4, -2]\). Therefore, \(f(-2)>f(-4)\).

a) Local minimum at \(x=-4\); stationary inflection point at \(x=-1\). b) Concave up on \([-5, -3)\) and \((-1, 0.5]\); concave down on \((-3, -1)\). c) \(x=-3\) and \(x=-1\) d) \(f(-2)>f(-4)\) because \(f\) is strictly increasing on \([-4, -2]\).

- Read the three marked points carefully. - At each marked local extremum, the first derivative is \(0\). - Use function-value and derivative conditions at extrema and inflection points. - The inflection point on the y-axis simplifies the cubic form. - A local extremum on the x-axis is a multiple zero.

1. From the graph, \(H=(-1, 4)\), \(T=(1, 0)\), and \(W=(0, 2)\). 2. Let \(f(x)=ax^3+bx^2+cx+d\). Since \(W=(0, 2)\) is an inflection point, \(f(0)=2\) and \(f''(0)=0\), so \(d=2\) and \(b=0\). 3. Using \(T=(1, 0)\), the equations \(f(1)=0\) and \(f'(1)=0\) give \(a+c=-2\) and \(3a+c=0\). Thus, \(a=1\) and \(c=-3\), so \(f(x)=x^3-3x+2\). 4. Factoring gives \(f(x)=(x-1)^2(x+2)\). Therefore, \(x=1\) is a double zero and \(x=-2\) is a simple zero.

a) \(H=(-1, 4)\), \(T=(1, 0)\), \(W=(0, 2)\) b) \(f(x)=x^3-3x+2\) c) \(x=1\), multiplicity \(2\); \(x=-2\), multiplicity \(1\)

- Read the exact coordinates of the labeled extrema from the graph. - At each extremum, use both the point and zero-slope conditions. - Substitute \(x=2\) to test Hana's point. - Find the midpoint of the two extrema.

1. From the graph, \(H=(-1, 4)\) and \(T=(1, 0)\). The extrema give \(f(-1)=4\), \(f(1)=0\), \(f'(-1)=0\), and \(f'(1)=0\). Solving these four equations gives \(a=1\), \(b=0\), \(c=-3\), and \(d=2\). Thus, \(f(x)=x^3-3x+2\). 2. Evaluating at \(x=2\) gives \(f(2)=8-6+2=4\ne5\), so \(P=(2, 5)\) is not on the graph. 3. The graph shows \(f(2)=4\) and \(W=(0, 2)\). The midpoint of \(H=(-1, 4)\) and \(T=(1, 0)\) is \((0, 2)\), so the extrema are paired by a half-turn about the inflection point.

a) \(f(x)=x^3-3x+2\), and \(f(2)=4\ne5\), so the claim is false. b) \(f(2)=4\) and \(W=(0, 2)\). The point \(W\) is the midpoint of \(H\) and \(T\), matching the graph's point symmetry.

- Compare turning points of one graph with zeros of the other. - Compare increasing and decreasing intervals with the sign of the candidate derivative graph. - Both relationships must agree for the match to be valid.

1. If one graph is \(f'\), its zeros must occur at horizontal tangents of \(f\). 2. Graph a has a local maximum at \(x=-2\) and a local minimum at \(x=2\). Graph b has zeros at exactly \(x=-2\) and \(x=2\). 3. Graph a decreases on \((-2,2)\), while graph b is negative there. Graph a increases for \(x<-2\) and \(x>2\), while graph b is positive there. 4. Therefore, graph a represents \(f\), and graph b represents \(f'\).

Graph a represents \(f\); graph b represents \(f'\).

- The sign of a derivative determines whether its original function increases or decreases. - A sign-changing derivative zero corresponds to a local extremum. - Match each candidate using both the critical numbers and the monotonicity pattern.

1. The graph of \(f'\) has zeros at \(x=0\) and \(x=2\). It is positive for \(x<0\) and \(x>2\), and negative for \(0<x<2\). 2. Therefore, \(f\) must increase, then decrease, then increase, with a local maximum at \(x=0\) and a local minimum at \(x=2\). Candidate a has this behavior. 3. The graph of \(g'\) is positive for \(x<1\), zero at \(x=1\), and negative for \(x>1\). 4. Therefore, \(g\) must increase before \(x=1\), decrease afterward, and have a local maximum at \(x=1\). Candidate b has this behavior.

\(f'\) matches candidate a for \(f\); \(g'\) matches candidate b for \(g\).

- Factor the function to find its zeros. - Compare \(f_a(-x)\) with \(f_a(x)\). - Use the first and second derivatives to locate and classify extrema. - Determine when \(\pm\sqrt{a}\) are distinct from zero. - For the final blueprint, distinguish the degenerate case \(a=0\) from \(a>0\); the derivative has a different zero structure in those two cases.

1. Factor \(f_a(x)=x(x^2-a)\). Graph p has only the zero \(x=0\), so \(a=0\). Graph q has zeros \(-1, 0, 1\), so \(a=1\). Graph r has zeros \(-2, 0, 2\), so \(a=4\). 2. Since \(f_a(-x)=-f_a(x)\), every graph is symmetric about the origin. 3. The first derivative is \(f_a'(x)=3x^2-a\), so the critical numbers are \(x=\pm\sqrt{\frac{a}{3}}\). The second derivative is \(f_a''(x)=6x\). 4. The local maximum is \(\left(-\sqrt{\frac{a}{3}}, \frac{2a}{3}\sqrt{\frac{a}{3}}\right)\), and the local minimum is \(\left(\sqrt{\frac{a}{3}}, -\frac{2a}{3}\sqrt{\frac{a}{3}}\right)\). 5. The zeros are \(0\) and \(\pm\sqrt{a}\). They are three distinct real numbers exactly when \(a>0\). 6. For \(a=0\), \(f_0(x)=x^3\): it is increasing on \(\mathbb{R}\), has no local extremum, has a stationary inflection point at \((0,0)\), and runs from \(-\infty\) to \(\infty\). 7. For \(a>0\), the graph increases on \(( -\infty,-\sqrt{a/3})\), decreases on \((-\sqrt{a/3},\sqrt{a/3})\), and increases on \((\sqrt{a/3},\infty)\). It has the local maximum and minimum from part c, a nonstationary inflection point at \((0,0)\), and the same cubic end behavior.

a) p: \(a=0\); q: \(a=1\); r: \(a=4\) b) Symmetric about the origin c) Local maximum: \(\left(-\sqrt{\frac{a}{3}}, \frac{2a}{3}\sqrt{\frac{a}{3}}\right)\); local minimum: \(\left(\sqrt{\frac{a}{3}}, -\frac{2a}{3}\sqrt{\frac{a}{3}}\right)\) d) \(a>0\) e) \(a=0\): increasing on \(\mathbb{R}\), no extrema, stationary inflection at \((0,0)\), cubic end behavior. For \(a>0\): increase → local maximum at \(x=-\sqrt{a/3}\) → decrease through the inflection at \(x=0\) → local minimum at \(x=\sqrt{a/3}\) → increase, with left end down and right end up.

- Use the sign of \(f'\) for monotonicity and local extrema. - Use whether \(f'\) is increasing or decreasing for concavity. - The comparison \(f(3)\) versus \(f(1)\) follows from monotonicity on \((1,3)\); no accumulation concept is needed.

1. Statement (1) is true. The graph of \(f'\) is positive on \((-2,1)\), so \(f\) is strictly increasing there. 2. Statement (2) is false. At \(x=1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 3. Statement (3) is true. The graph of \(f'\) has a local minimum at approximately \(x=2.1\), changing from decreasing to increasing; therefore \(f\) changes from concave down to concave up there. 4. Statement (4) is false. Since \(f'(x)<0\) on \((1,3)\), \(f\) is strictly decreasing on that interval, so \(f(3)<f(1)\).

(1) True. (2) False. (3) True. (4) False.

- A derivative's zeros correspond to horizontal tangents of the function above it in the chain. - The sign of a derivative must agree with the increasing/decreasing behavior of its parent function. - Test the graphs pairwise until two consecutive derivative relationships both work.

1. Graph b has a local maximum at \(x=-1\) and a local minimum at \(x=1\). Graph a has zeros at those x-values and signs consistent with the increasing and decreasing intervals of graph b. Thus graph a is the derivative of graph b. 2. Graph c has extrema where graph b is zero, and its monotonicity agrees with the sign of graph b. Thus graph b is the derivative of graph c. 3. Therefore, graph c represents \(f\), graph b represents \(f'\), and graph a represents \(f''\). 4. Graph d has the opposite monotonicity relationship and cannot complete this derivative chain.

Graph c is \(f\); graph b is \(f'\); graph a is \(f''\); graph d does not belong.

- Use the sign of the derivative graph for monotonicity and extrema. - Verify the candidate function by differentiating it. - For concavity, track whether the derivative graph is increasing or decreasing.

1. Since \(f'>0\) on \((-2,2)\), \(f\) is strictly increasing there. Since \(f'<0\) on \(( -\infty,-2)\) and \((2,\infty)\), \(f\) is strictly decreasing there. The sign changes give a local minimum at \(x=-2\) and a local maximum at \(x=2\). 2. Differentiating \(f(x)=-\frac13x^3+4x-1\) gives \(f'(x)=4-x^2\), matching the displayed graph. 3. Yes. The derivative graph has a local maximum at \(x=0\), changing from increasing to decreasing. Therefore \(f\) changes from concave up to concave down at \(x=0\).

a) Increasing on \((-2,2)\); decreasing on \(( -\infty,-2)\) and \((2,\infty)\). Local minimum at \(x=-2\); local maximum at \(x=2\). b) \(f'(x)=4-x^2\), matching the graph. c) Yes, \(x=0\) is an inflection point.

- Set \(f'(x) = 0\) by finding where \(g(x) = 1\). - Compare \(g(x)\) with \(1\) on each interval. - An inflection point of \(f\) occurs where \(f'\) has a local extremum.

1. Local extrema of \(f\) occur where \(f'(x) = 0\), so solve \(g(x) = 1\). From the graph, this occurs at \(x = -2\) and \(x = 2\). 2. For \(x < -2\), \(g(x) < 1\), so \(f'(x) < 0\). For \(-2 < x < 2\), \(g(x) > 1\), so \(f'(x) > 0\). Thus, \(f\) has a local minimum at \(x = -2\). 3. For \(-2 < x < 2\), \(f'(x) > 0\), and for \(x > 2\), \(f'(x) < 0\). Thus, \(f\) has a local maximum at \(x = 2\). 4. Since \(f''(x) = g'(x)\), an inflection point of \(f\) occurs where \(g\) has a local extremum. The graph of \(g\) has a maximum at \(x = 0\), so \(f\) has an inflection point there.

a) Local minimum at \(x = -2\); local maximum at \(x = 2\) b) Inflection point at \(x = 0\)

- Use zeros and sign changes of \(f'\) to classify local extrema. - The largest value of \(f'\) is the greatest slope of \(f\). - Parallel lines have equal slopes. - Read the vertex to identify \(d\) and \(c\), then use another point to find \(a\).

1. The derivative is zero at \(x = 0\) and \(x = 4\). At \(x = 0\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x = 4\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 2. The greatest slope of \(f\) is the maximum value of \(f'\). The vertex of the derivative graph is \((2, 2)\), so the greatest slope is \(2\) at \(x = 2\). 3. A line parallel to \(g\) must have slope \(1.5\). The graph shows \(f'(x) = 1.5\) at \(x = 1\) and \(x = 3\). 4. The vertex gives \(d = 2\) and \(c = 2\). Using \((0, 0)\), \(0 = a(0-2)^2 + 2\), so \(4a = -2\) and \(a = -0.5\).

a) Local minimum at \(x = 0\); local maximum at \(x = 4\) b) Greatest slope \(2\) at \(x = 2\) c) \(x = 1\) and \(x = 3\) d) \(a = -0.5\), \(d = 2\), and \(c = 2\)

- Compare zeros and extrema to test which graph could be the derivative of another. - The slope of \(f\) is read from \(f'\). - Compare numerator and denominator degrees for the end behavior of \(f''\).

1. Graph a represents \(f\). Its derivative is \(f'(x)=\frac{3x}{x^2+1}\), which is graph b. Differentiating again gives \(f''(x)=\frac{3-3x^2}{(x^2+1)^2}\), which is graph c. 2. The slope of \(f\) at \(x=2\) is \(f'(2)=\frac65\). 3. The denominator of \(f''\) has higher degree than its numerator, so \(f''(x)\to0\) as \(x\to\pm\infty\). The horizontal asymptote is \(y=0\).

a) a: \(f\); b: \(f'\); c: \(f''\). b) \(\frac65\). c) \(\lim_{x\to\pm\infty}f''(x)=0\); horizontal asymptote \(y=0\).

- Read where the derivative graph is above and below the x-axis. - Translate positive derivative into increasing behavior and negative derivative into decreasing behavior. - Look for sign changes of the derivative before comparing the candidate function graphs.

1. In panel a), \(f'(x)>0\) for \(-2.5<x<-1\), \(f'(x)<0\) for \(-1<x<1\), and \(f'(x)>0\) for \(1<x<2.5\). 2. Therefore \(f\) must increase, then decrease, then increase across those three intervals. 3. Candidate b) has exactly that monotonicity pattern. Candidate c) has the opposite pattern, and candidate d) is increasing throughout the displayed interval. 4. Because \(f'\) changes from positive to negative at \(x=-1\), \(f\) has a local maximum there. Because \(f'\) changes from negative to positive at \(x=1\), \(f\) has a local minimum there.

Panel b) is consistent with \(f'\). \(f\) is increasing on \((-2.5,-1)\), decreasing on \((-1,1)\), and increasing on \((1,2.5)\). Local maximum x-coordinate: \(-1\). Local minimum x-coordinate: \(1\).

- Use the sign of \(f'\) for monotonicity and local extrema. - Use whether \(f'\) is increasing or decreasing for concavity. - Count changes in the direction of \(f'\), not merely zeros of \(f'\), when finding inflection points.

1. On \((-2,2)\), \(f'(x)<0\), so \(f\) is strictly decreasing. Statement (1) is true. 2. At \(x=2\), \(f'\) changes from negative to positive, so \(f\) has a local minimum, not a local maximum. Statement (2) is false. 3. On \((0,3)\), the graph of \(f'\) is increasing, so \(f\) is concave up. Statement (3) is true. 4. Inflection points of \(f\) occur where \(f'\) changes between decreasing and increasing. The parabola \(f'\) has only one extremum, at \(x=0\), so \(f\) has exactly one inflection point. Statement (4) is false.

(1) True. (2) False. (3) True. (4) False.

- Use the sign of \(f'\) to determine increasing and decreasing intervals. - Use the direction of each sign change to classify local extrema. - A local extremum of \(f'\) corresponds to a possible inflection point of \(f\).

1. The derivative is negative on \((-4, -1)\) and \((3, 6)\), so \(f\) is strictly decreasing on \([-4, -1]\) and \([3, 6]\). It is positive on \((-1, 3)\), so \(f\) is strictly increasing on \([-1, 3]\). 2. At \(x = -1\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x = 3\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 3. The graph of \(f'\) has a local maximum at \(x = 1\). Therefore, \(f''\) changes sign there, so \(f\) changes concavity and has an inflection point at \(x = 1\).

a) Increasing on \([-1, 3]\); decreasing on \([-4, -1]\) and \([3, 6]\) b) Local minimum at \(x = -1\); local maximum at \(x = 3\) c) Yes; the inflection point occurs at \(x = 1\).

- Use derivative sign changes to classify the two critical numbers. - Verify the candidate by differentiating rather than reconstructing it from the derivative. - Substitute \(x=3\) only after the candidate has been verified.

1. At \(x=0\), \(g'\) changes from positive to negative, so \(g\) has a local maximum. At \(x=2\), \(g'\) changes from negative to positive, so \(g\) has a local minimum. 2. Differentiating the candidate gives \(g'(x)=1.5x^2-3x\), which has zeros \(0\) and \(2\) and vertex \((1,-1.5)\), matching the graph. 3. \(g(3)=0.5(27)-1.5(9)+2=2\).

a) Local maximum at \(x=0\); local minimum at \(x=2\). b) \(g'(x)=1.5x^2-3x\), matching the graph, and \(g(3)=2\).

- Calculate each y-intercept and find the zeros. - Differentiate each function and use the derivative sign to locate extrema. - Compare end behavior in both directions. - Match the resulting features with the displayed graphs.

1. For \(f(x)=(x-2)e^x\), \(f(0)=-2\), and the only zero is \(x=2\). Also, \(f'(x)=(x-1)e^x\), so \(f\) decreases for \(x<1\), increases for \(x>1\), and has a local minimum at \(x=1\). Since \(f(x)\to0\) as \(x\to-\infty\), these features match graph \(A\). 2. For \(g(x)=3-e^x\), \(g(0)=2\), \(g'(x)=-e^x<0\), so the function is strictly decreasing, and \(g(x)\to3\) as \(x\to-\infty\). These features match graph \(B\). 3. For \(h(x)=xe^{-x}\), the graph passes through \((0, 0)\). Since \(h'(x)=(1-x)e^{-x}\), it increases for \(x<1\), decreases for \(x>1\), and has a local maximum at \(x=1\). Also, \(h(x)\to0\) as \(x\to\infty\). These features match graph \(C\).

Graph \(A\): function 1; graph \(B\): function 2; graph \(C\): function 3.

- Read the coordinates of the marked local maximum and local minimum. - Count the turning points shown in the graph. - Relate the maximum possible number of turning points to polynomial degree. - Use origin symmetry to determine which powers may appear. - At the marked local minimum, use both the point and zero-slope conditions.

1. The graph has two local extrema. A degree-\(n\) polynomial can have at most \(n-1\) local extrema, so the least possible degree is \(3\). 2. The graph is symmetric about the origin, so a cubic of the least possible degree has the form \(f(x)=ax^3+cx\). 3. Using the local minimum \(T=(2,-4)\), the conditions \(f(2)=-4\) and \(f'(2)=0\) give \(4a+c=-2\) and \(12a+c=0\). Solving gives \(a=\frac14\) and \(c=-3\). Therefore, \(f(x)=\frac14x^3-3x\).

a) Degree \(3\) b) \(f(x)=ax^3+cx\) c) \(f(x)=\frac{1}{4}x^3-3x\)

- Locate the horizontal tangent lines on the graph of \(f\). - The derivative must be zero at those inputs. - Determine where \(f\) is increasing and decreasing, then match the derivative sign. - A cubic polynomial has a quadratic derivative.

1. The graph of \(f\) has a local maximum near \(x=-0.15\) and a local minimum near \(x=2.15\). Therefore, \(f^{\prime}\) must have zeros near those inputs. Both \(g\) and \(h\) have the required zeros. 2. Between the two extrema, \(f\) is decreasing. Therefore, \(f^{\prime}\) must be negative on that interval. 3. Graph \(g\) lies below the x-axis between the two zeros, while graph \(h\) lies above it. Thus, \(g\) has the correct sign pattern. 4. Graph \(k\) has only one zero and is linear, so it cannot be the derivative of the displayed cubic function. Therefore, graph \(g\) represents \(f^{\prime}\).

The red graph \(g\) represents \(f^{\prime}\).

- Zeros of the derivative correspond to horizontal tangents of the original function. - Where \(f'>0\), \(f\) increases; where \(f'<0\), it decreases. - Eliminate candidates that fail either condition.

1. The derivative graph \(f'\) has zeros at \(x=-2\) and \(x=2\), so \(f\) must have horizontal tangents at those x-values. Graphs \(g\) and \(h\) do, while graph \(k\) does not. 2. On \((-2,2)\), \(f'(x)<0\), so \(f\) must be decreasing there. Graph \(g\) decreases on that interval, while graph \(h\) increases. 3. Therefore, graph \(g\) represents \(f\).

Graph \(g\) represents \(f\).

- Identify the graph roles before reading any requested values. - Zeros of a derivative correspond to horizontal tangents of its function. - Once the derivative graph is identified, its y-values give tangent slopes of the function.

1. Figure 1 is \(g\), and Figure 2 is \(g'\). Figure 1 has horizontal tangents at \(x=-2,0,2\), exactly where Figure 2 crosses the x-axis. 2. Figure 1 gives \(g(2)=-3\). 3. The slope at \(x=2\) is \(g'(2)=0\). 4. Figure 2 gives \(g'(1)=-3\). 5. Figure 1 shows \(g(0)=1\), so \(x=0\) works. 6. Horizontal tangents occur where \(g'(x)=0\), at \(x=-2,0,2\). 7. Figure 2 shows \(g'(-1)=3\), so \(x=-1\) works.

a) Figure 1 is \(g\); Figure 2 is \(g'\). b) \(g(2)=-3\) c) \(0\) d) \(-3\) e) \(x=0\) f) \(x=-2,0,2\) g) \(x=-1\)

- Locate the local maximum and local minimum of \(f\). - The derivative is zero at horizontal tangent lines. - Look for the point where the graph rises most steeply.

1. The derivative is zero at the local extrema of \(f\). The graph has a local minimum at \(x=-\frac{\pi}{2}\) and a local maximum at \(x=\frac{\pi}{2}\). Therefore, the graph of \(f'\) crosses the x-axis at those two inputs. 2. The derivative is greatest where the graph of \(f\) rises most steeply. This occurs at the origin, so \(f'\) reaches its greatest value at \(x=0\).

1) \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\) 2) \(x=0\)

- A smooth local maximum or minimum occurs where the derivative is zero. - Read \(p(2)\) from the graph of \(p\) and \(p'(2)\) from the derivative graph.

1. The graph of \(p\) has a local maximum at \(x = -1\) and a local minimum at \(x = 3\). The graph of \(q\) is zero at those same \(x\)-values. Therefore, \(q = p'\). 2. Reading from the graph of \(p\), \(p(2) \approx -0.4\). 3. Since \(q = p'\), \(p'(2) = q(2) \approx -1.8\).

a) \(q\) is the derivative of \(p\), so \(q = p'\). b) \(p(2) \approx -0.4\) and \(p'(2) \approx -1.8\)

- The derivative gives the tangent slope of \(f\). - Local extrema of \(f\) produce zeros of \(f^{\prime}\). - Increasing and decreasing intervals determine the sign of the derivative. - Use the symmetry of the original graph to infer the symmetry of its derivative.

1. Statement (1) is true. The graph of \(f\) is increasing at \(x=0\), so \(f^{\prime}(0)>0\). 2. Statement (2) is true. The graph has two local extrema, near \(x=-1.4\) and \(x=1.4\). The tangent line is horizontal at each point, so \(f^{\prime}\) has exactly two zeros. 3. Statement (3) is true. The graph of \(f\) is decreasing on \([1.5,3]\), so \(f^{\prime}(x)<0\) throughout that interval. 4. Statement (4) is true. The graph of \(f\) has point symmetry about \((0,1)\), so its derivative is symmetric about the y-axis. Therefore, \(f^{\prime}(2)=f^{\prime}(-2)\). 5. Statement (5) is false. Passing through the origin would require \(f^{\prime}(0)=0\), but the graph of \(f\) has positive slope at \(x=0\).

(1) True (2) True (3) True (4) True (5) False

- Find the derivative from the given formula to check statements extending beyond the graph. - Examine the signs of its numerator and denominator separately. - Test what happens when \(x\) is replaced by \(-x\).

1. By the quotient rule, \(g'(x)=\frac{-8x}{(x^2+1)^2}\). 2. Statement (1) is true: the graph has a smooth maximum at \(0\), and \(g'(0)=0\). 3. Statement (2) is true: for \(x<0\), the numerator \(-8x\) is positive and the denominator is positive. 4. Statement (3) is false: the derivative has exactly one zero, at \(x=0\). 5. Statement (4) is true: the graph falls near \(1\), and \(g'(1)=-2<0\). 6. Statement (5) is true: \(g'(-x)=-g'(x)\), so its graph is symmetric about the origin.

(1) True; horizontal tangent at \(x=0\). (2) True; the derivative formula is positive for \(x<0\). (3) False; the pictured curve has only one horizontal tangent in the interval, at \(x=0\), and the derivative formula confirms it is the only zero. (4) True; graph falls near \(1\), where \(g'(1)=-2\). (5) True; \(g'(-x)=-g'(x)\).

- Horizontal tangent lines on \(f\) give zeros of \(f^{\prime}\). - Determine whether \(f\) increases or decreases on each interval between its extrema. - The derivative of a fourth-degree polynomial is cubic. - Use the symmetry of \(f\) to check the symmetry of \(f^{\prime}\).

1. The graph of \(f\) has local maxima at \(x=-2\) and \(x=2\) and a local minimum at \(x=0\). Therefore, \(f^{\prime}(-2)=0\), \(f^{\prime}(0)=0\), and \(f^{\prime}(2)=0\). 2. The function increases for \(x<-2\), decreases for \(-2<x<0\), increases for \(0<x<2\), and decreases for \(x>2\). Thus, \(f^{\prime}\) has the sign pattern positive, negative, positive, negative across those intervals. 3. Since \(f\) is a fourth-degree polynomial, \(f^{\prime}\) is cubic. It is an origin-symmetric cubic that crosses the x-axis at \(x=-2\), \(x=0\), and \(x=2\) with the sign pattern positive, negative, positive, negative from left to right.

a) \(x=-2\), \(x=0\), and \(x=2\) b) An origin-symmetric cubic crossing the x-axis at \(x=-2\), \(x=0\), and \(x=2\), with signs \(+,-,+,-\) from left to right.

- Local extrema of \(f'\) correspond to inflection points of \(f\). - The monotonicity of \(f'\) determines the concavity of \(f\). - What does a positive derivative imply when comparing two function values? - Use the sign change of \(f'\) to classify a critical number of \(f\).

1. Statement 1 is true. The graph of \(f'\) has two local extrema on \([-3, 3]\), so \(f\) has two inflection points. 2. Statement 2 is false. The derivative \(f'\) decreases only until \(x=\frac{2}{\sqrt{3}}\approx1.15\), then increases. Therefore, \(f\) is not concave down on the entire interval \([0, 2]\). 3. Statement 3 is true. Since \(f'(x)>0\) for \(2<x<3\), the function \(f\) is increasing there. Therefore, \(f(2)<f(3)\). 4. Statement 4 is false. At \(x=0\), \(f'\) changes from positive to negative, so \(f\) has a local maximum, not a local minimum.

1) True 2) False 3) True 4) False

- Use the sign of \(f'\) to determine whether \(f\) is increasing or decreasing. - Use sign changes of \(f'\) to classify critical points of \(f\). - The monotonicity of \(f'\) determines the concavity of \(f\).

1. Statement a is true. On \((1, 3)\), \(f'(x)<0\), with zeros at the endpoints. Therefore, \(f\) is strictly decreasing on \([1, 3]\). 2. Statement b is true. At \(x=1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 3. Statement c is false. At \(x=0\), the graph of \(f'\) is decreasing, so \(f''(0)<0\). Therefore, \(f\) is concave down there.

a) True b) True c) False

- Factor the polynomial to find its zeros. - Use the first and second derivatives to find the local maximum. - Set the second derivative equal to \(0\) for possible inflection points. - Use the graph's symmetry to check paired answers.

1. Since \(f(x)=\frac{1}{4}(x^2-4)^2\), the zeros are \(x=-2\) and \(x=2\), each with multiplicity \(2\). Thus, \(N_1=(-2, 0)\) and \(N_2=(2, 0)\). 2. The derivatives are \(f'(x)=x^3-4x\), \(f''(x)=3x^2-4\), and \(f'''(x)=6x\). The second-derivative test gives a local maximum at \(H=(0, 4)\). 3. Solving \(f''(x)=0\) gives \(x=\pm\frac{2\sqrt{3}}{3}\). At either value, \(f(x)=\frac{16}{9}\), and the third derivative is nonzero. Therefore, \(W_1=(-\frac{2\sqrt{3}}{3}, \frac{16}{9})\) and \(W_2=(\frac{2\sqrt{3}}{3}, \frac{16}{9})\).

\(N_1=(-2, 0)\), \(N_2=(2, 0)\), \(H=(0, 4)\), \(W_1=(-\frac{2\sqrt{3}}{3}, \frac{16}{9})\), and \(W_2=(\frac{2\sqrt{3}}{3}, \frac{16}{9})\)

- Factor the polynomial to find zeros and multiplicities. - Use the first and second derivatives for extrema. - Use the second derivative for the inflection point. - Compare each coordinate with the graph.

1. Factoring gives \(f(x)=x(x-3)^2\). Therefore, \(x=0\) is a simple zero and \(x=3\) is a double zero. 2. The derivatives are \(f'(x)=3(x-1)(x-3)\), \(f''(x)=6x-12\), and \(f'''(x)=6\). The second-derivative test gives a local maximum at \(H=(1, 4)\) and a local minimum at \(T=(3, 0)\). 3. Solving \(f''(x)=0\) gives \(x=2\), and \(f(2)=2\). Thus, \(W=(2, 2)\). All points agree with the graph.

Zeros: \((0, 0)\), simple, and \((3, 0)\), double; local maximum: \(H=(1, 4)\); local minimum: \(T=(3, 0)\); inflection point: \(W=(2, 2)\) All points agree with the graph.

- Use zeros and sign changes of \(f'\) to identify local extrema. - The function decreases where its derivative is negative. - Inflection points of \(f\) correspond to local extrema of \(f'\).

1. The derivative is zero at \(x = -3\), \(x = 1\), and \(x = 5\). 2. At \(x = -3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x = 1\), it changes from positive to negative, so \(f\) has a local maximum. At \(x = 5\), it changes from negative to positive, so \(f\) has a local minimum. 3. The derivative is negative on \((-4, -3)\) and \((1, 5)\), so \(f\) is strictly decreasing on \([-4, -3]\) and \([1, 5]\). 4. Inflection points of \(f\) occur where \(f'\) has local extrema. The derivative graph has exactly two local extrema in the displayed interval, so \(f\) has two inflection points.

a) Local minima at \(x = -3\) and \(x = 5\); local maximum at \(x = 1\) b) \([-4, -3]\) and \([1, 5]\) c) Two inflection points

- Compare the general shapes of a cubic function and its derivative. - The derivative is zero where the original graph has horizontal tangents. - Adding a constant outside a function shifts its graph vertically. - Replacing \(x\) with \(x-c\) shifts a graph horizontally.

1. Differentiate: \(f'(x)=x^2-1\). This is an upward-opening parabola with vertex \((0, -1)\) and zeros at \(x=-1\) and \(x=1\), so it is graph I. 2. The original function \(f\) is the cubic with a local maximum at \((-1, \frac{2}{3})\) and a local minimum at \((1, -\frac{2}{3})\), so it is graph II. 3. The function \(h(x)=f(x-2)\) shifts the graph of \(f\) right \(2\) units, so it is graph III. 4. The function \(g(x)=f(x)+2\) shifts the graph of \(f\) up \(2\) units, so it is graph IV.

I: \(f^{\prime}(x)\); II: \(f(x)\); III: \(h(x)=f(x-2)\); IV: \(g(x)=f(x)+2\).

- Use zeros and sign changes of \(f'\) to identify local extrema. - The function decreases where \(f' < 0\). - A local extremum of \(f'\) indicates a possible inflection point of \(f\).

1. At \(x = 0\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. Statement a) is true. 2. For \(0 < x < 4\), \(f'(x) < 0\), so \(f\) is strictly decreasing on \([0, 4]\). Statement b) is true. 3. The graph of \(f'\) has a local minimum at \(x = 2\), so \(f''\) changes sign there. Therefore, \(f\) changes concavity and has an inflection point at \(x = 2\). Statement c) is true.

a) True b) True c) True

- The monotonicity of \(f'\) determines the concavity of \(f\). - Use sign changes of \(f'\) to identify local extrema of \(f\). - Local extrema of \(f'\) correspond to inflection points of \(f\). - Use the sign of \(f'\) to determine whether \(f\) increases or decreases.

1. Statement a is false. For \(x>2\), the graph of \(f'\) is decreasing, so \(f''(x)<0\) and \(f\) is concave down. 2. Statement b is true. At \(x=4\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 3. Statement c is true. The graph of \(f'\) has a local maximum at \(x=2\), so the concavity of \(f\) changes there. 4. Statement d is false. For \(0<x<2\), \(f'(x)>0\), so \(f\) is strictly increasing there.

a) False b) True c) True d) False

- Zeros and sign changes of \(f^{\prime}\) determine local extrema of \(f\). - The function \(f\) decreases where \(f^{\prime}<0\). - Inflection points of \(f\) occur where \(f^{\prime}\) changes monotonicity.

1. Local extrema of \(f\) occur where \(f^{\prime}\) changes sign. The graph has zeros at approximately \(x=-2.4\), \(x=0\), and \(x=2.4\). The sign changes from negative to positive at \(x\approx-2.4\), so \(f\) has a local minimum there. It changes from positive to negative at \(x=0\), so \(f\) has a local maximum there. It changes from negative to positive at \(x\approx2.4\), so \(f\) has a local minimum there. 2. The function \(f\) is strictly decreasing where \(f^{\prime}<0\), which on the displayed interval is approximately \((-4, -2.4)\) and \((0, 2.4)\). 3. Inflection points of \(f\) occur where \(f^{\prime}\) changes from increasing to decreasing or from decreasing to increasing. The graph of \(f^{\prime}\) has local extrema at approximately \(x=-1.4\) and \(x=1.4\), so those are the inflection-point x-values of \(f\).

1. Local minima at \(x\approx-2.4\) and \(x\approx2.4\); local maximum at \(x=0\) 2. Approximately \((-4, -2.4)\) and \((0, 2.4)\) 3. Approximately \(x=-1.4\) and \(x=1.4\)

- Match local extrema with zeros of the derivative. - Compare increasing and decreasing intervals with the sign of the derivative. - Use the degree and symmetry of each graph as a check.

1. Graph (1) has local minima at \(x=-2\) and \(x=2\) and a local maximum at \(x=0\). Its derivative must be zero at \(-2\), \(0\), and \(2\), so (1) matches (B). 2. Graph (2) is \(2\sin x\). Its derivative is \(2\cos x\), so (2) matches (C). 3. Graph (3) is a downward-opening parabola with vertex at \(x=0\). Its derivative is a decreasing line through the origin, so (3) matches (A).

(1) \(\rightarrow\) (B); (2) \(\rightarrow\) (C); (3) \(\rightarrow\) (A).

- Match local extrema with zeros of the derivative. - Compare increasing and decreasing behavior with the sign of the derivative. - Consider behavior near asymptotes and at the edges of the displayed interval.

1. Graph (1) is \(e^x - 2\), which is increasing everywhere. Its derivative is \(e^x\), which is always positive, so (1) matches (A). 2. Graph (2) is \(\frac{1}{x}\), which decreases on both parts of its domain. Its derivative is \(-\frac{1}{x^2}\), which is always negative, so (2) matches (C). 3. Graph (3) has a local maximum at \(x = -1\) and a local minimum at \(x = 1\). Its derivative must be an upward-opening parabola with zeros at \(x = -1\) and \(x = 1\), so (3) matches (B).

(1) \(\rightarrow\) (A); (2) \(\rightarrow\) (C); (3) \(\rightarrow\) (B).

- Factor the polynomial to find x-intercepts and multiplicities. - Use the first and second derivatives to find and classify extrema. - Use the second derivative to find the inflection point. - Compare the exact coordinates with the graph.

1. Since \(f(x)=x^2(x-3)\), the x-intercepts are \((0, 0)\), with multiplicity \(2\), and \((3, 0)\), with multiplicity \(1\). The y-intercept is also \((0, 0)\). 2. The derivatives are \(f'(x)=3x(x-2)\), \(f''(x)=6x-6\), and \(f'''(x)=6\). The second-derivative test gives a local maximum at \((0, 0)\) and a local minimum at \((2, -4)\). 3. The inflection point occurs at \(x=1\), where \(f(1)=-2\). Thus, the inflection point is \((1, -2)\). 4. These exact points agree with the diagram, so the graph is correct.

Yes, the graph is correct. Intercepts: \((0, 0)\), a double x-intercept and the y-intercept; \((3, 0)\), a simple x-intercept Local maximum: \((0, 0)\); local minimum: \((2, -4)\) Inflection point: \((1, -2)\)

- Evaluate the function at \(x=0\). - Factor the cubic to find zeros and multiplicities. - Use the first and second derivatives to find and classify extrema. - Compare the exact values with the graph's scale.

1. The y-intercept is \((0, 2)\). Factoring gives \(f(x)=-(x+1)^2(x-2)\), so the x-intercepts are \((-1, 0)\), with multiplicity \(2\), and \((2, 0)\), with multiplicity \(1\). 2. The derivatives are \(f'(x)=-3x^2+3\) and \(f''(x)=-6x\). The critical numbers are \(x=-1\) and \(x=1\). The second-derivative test gives a local minimum at \((-1, 0)\) and a local maximum at \((1, 4)\). 3. The displayed graph has y-intercept \((0, 1.6)\) and a local maximum near \((1, 3.2)\). These do not match the exact values, so the graph is incorrect.

No. Intercepts: \((0, 2)\), \((-1, 0)\) with multiplicity \(2\), and \((2, 0)\) Local minimum: \((-1, 0)\); local maximum: \((1, 4)\)

- Check \(f(0)\) before doing further calculations. - Find the zeros of the first derivative; if the first and second derivatives are both \(0\), check the third derivative. - Compute the exact stationary-point coordinates and compare them with the graph.

1. The y-intercept is \((0, 2)\), but the displayed graph has y-intercept \((0, 4)\). This already shows that the graph is incorrect. 2. The derivative is \(f'(x)=x^2(x-3)\). Thus, the critical numbers are \(x=0\) and \(x=3\). 3. At \(x=3\), \(f''(3)=9>0\), so there is a local minimum at \((3, -\frac{19}{4})\). At \(x=0\), the first and second derivatives are \(0\), while \(f'''(0)=-6\ne0\), so \((0, 2)\) is a stationary inflection point. 4. These points do not match the displayed graph, so the graph is incorrect.

No. Y-intercept and stationary inflection point: \((0, 2)\) Local minimum: \((3, -\frac{19}{4})\)

- Find the critical numbers from the first derivative and classify them with the second derivative. - A local minimum on the x-axis is a multiple zero. - Set the second derivative equal to \(0\) to locate the inflection point. - Compare the exact critical numbers and inflection-point location with the graph.

1. The y-intercept is \((0,4)\). The derivative is \(f'(x)=3x(x-2)\), so the critical numbers are \(x=0\) and \(x=2\). 2. Since \(f''(0)=-6<0\), \((0,4)\) is a local maximum. Since \(f''(2)=6>0\), \((2,0)\) is a local minimum. 3. Because the local minimum lies on the x-axis, \(x=2\) is a double zero. Factoring gives \(f(x)=(x-2)^2(x+1)\), so the other x-intercept is \((-1,0)\). 4. Solving \(f''(x)=6x-6=0\) gives \(x=1\). Since \(f(1)=2\), the inflection point is \((1,2)\). 5. The displayed graph places the local minimum at \(x=3\), not \(x=2\), and its inflection point is also misplaced. Therefore, it is incorrect.

No. Intercepts: \((0,4)\), \((-1,0)\), and \((2,0)\), where \(x=2\) is a double zero Local maximum: \((0,4)\); local minimum: \((2,0)\) Inflection point: \((1,2)\)

- Match local extrema of a function with zeros of its derivative. - Compare increasing and decreasing intervals with the sign of the derivative. - Use the function types and polynomial degrees as a check.

1. The function \(f_1\) matches \(g_1\), and \(f_2\) matches \(g_2\). 2. The graph of \(f_1\) has a local maximum at \(x = -2\) and a local minimum at \(x = 2\). Graph \(g_1\) is zero at those values. Between them, \(f_1\) decreases, and \(g_1\) is negative. 3. Therefore, \(f_1\) is strictly decreasing on \([-2, 2]\).

1. \(f_1 \rightarrow g_1\); \(f_2 \rightarrow g_2\) 2. The zeros and sign of \(g_1\) match the horizontal tangents and monotonicity of \(f_1\). 3. \([-2, 2]\)

- Compare the polynomial degrees. - Match zeros of the derivative with horizontal tangents of the function. - The \(x\)-coordinate of an inflection point of \(f\) corresponds to a local extremum of \(f'\).

1. The blue graph is cubic and the red graph is quadratic, consistent with differentiation lowering the degree by one. Also, the zeros of the red graph occur at the horizontal tangents of the blue graph. Therefore, blue represents \(f\), and red represents \(f'\). 2. The derivative is zero at \(x = -1\) and \(x = 2\). At \(x = -1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. At \(x = 2\), it changes from negative to positive, so \(f\) has a local minimum. 3. The vertex of the derivative graph is at \(x = 0.5\), so \(f\) changes concavity there. Reading the blue graph to the nearest tenth gives \(f(0.5) \approx 0.4\), so the inflection point is approximately \((0.5, 0.4)\).

a) Blue represents \(f\), and red represents \(f'\). b) Local maximum at \(x = -1\); local minimum at \(x = 2\) c) Approximately \((0.5, 0.4)\)

- Count the x-axis crossings. - Use the polynomial degree to bound the number of zeros. - Identify the graph's local maximum and local minimum. - Use the sign and degree of the leading term for end behavior. - Relate the two extrema to the shape of the quadratic derivative.

1. The graph crosses the x-axis three times, so \(f\) has three real zeros. A degree-\(3\) polynomial can have at most three real zeros because a nonzero polynomial cannot have more zeros than its degree. 2. The graph has two local extrema. Since the leading term is \(-\frac{1}{2}x^3\), \(f(x)\to-\infty\) as \(x\to\infty\), and \(f(x)\to\infty\) as \(x\to-\infty\). 3. The derivative is a quadratic with two distinct zeros at the extrema. Its vertex lies between those zeros, where the derivative changes from decreasing to increasing or vice versa. Therefore, the second derivative changes sign there, producing an inflection point between the extrema.

a) Three real zeros; a cubic has at most three real zeros. b) Two local extrema; \(f(x)\to-\infty\) as \(x\to\infty\), and \(f(x)\to\infty\) as \(x\to-\infty\). c) The quadratic derivative has its vertex between its two zeros, so the second derivative changes sign there.

- Zeros of \(f^{\prime}\) give locations of horizontal tangents on \(f\). - Where \(f^{\prime}>0\), \(f\) increases. - Where \(f^{\prime}<0\), \(f\) decreases. - Where \(f^{\prime}\) changes from increasing to decreasing, \(f\) changes concavity.

1. The derivative is \(0\) at \(x=-2\) and \(x=4\). It changes from negative to positive at \(x=-2\), so a possible graph of \(f\) has a local minimum there. It changes from positive to negative at \(x=4\), so the graph has a local maximum there. The graph of \(f^{\prime}\) increases up to \(x=1\) and decreases after \(x=1\), so \(f\) changes from concave up to concave down and has an inflection point at \(x=1\). Any vertical shift with these slope and concavity properties is acceptable. 2. Since \(f^{\prime}<0\) on \((-3, -2)\), \(f\) is decreasing there. Since \(f^{\prime}>0\) on \((-2, 4)\), \(f\) is increasing there. Since \(f^{\prime}<0\) on \((4, 5)\), \(f\) is decreasing there.

1. One possible graph has a local minimum at \(x=-2\), an inflection point at \(x=1\), and a local maximum at \(x=4\), with slopes matching the displayed derivative. 2. Decreasing on \((-3, -2)\) and \((4, 5)\); increasing on \((-2, 4)\)

- A zero of the derivative without a sign change does not produce a local extremum. - Use whether \(f^{\prime}\) is increasing or decreasing to determine the concavity of \(f\). - Determine whether \(f^{\prime}\) is increasing or decreasing on each side of \(x=3\). - A horizontal tangent together with a change in concavity indicates a stationary inflection point.

1. The derivative changes from negative to positive at \(x=0\), so a possible graph of \(f\) has a local minimum at \(x=0\). The graph of \(f^{\prime}\) increases up to \(x=1\), decreases from \(x=1\) to \(x=3\), and increases after \(x=3\). Therefore, \(f\) is concave up, then concave down, then concave up, with inflection points at \(x=1\) and \(x=3\). At \(x=3\), \(f^{\prime}=0\), but \(f^{\prime}\) is positive on both sides, so \(f\) remains increasing and has no local extremum there. 2. The graph of \(f^{\prime}\) has a local minimum at \(x=3\). Therefore, \(f^{\prime}\) changes from decreasing to increasing, so the concavity of \(f\) changes. Because \(f^{\prime}(3)=0\), \(f\) has a stationary inflection point at \(x=3\).

1. One possible graph has a local minimum at \(x=0\), an inflection point at \(x=1\), and a stationary inflection point at \(x=3\); it remains increasing through \(x=3\). 2. At \(x=3\), \(f\) has a stationary inflection point: its tangent is horizontal, but the function remains increasing through the point.

- The sign of \(f'\) gives the direction of the original function. - A zero of \(f'\) that merely touches the x-axis behaves differently from a sign-changing zero. - Local extrema of \(f'\) correspond to changes in the concavity of \(f\). - Use the monotonicity on the full displayed interval to determine where the greatest value occurs.

1. The graph of \(f'\) is positive on \((-3,0)\) except that it touches \(0\) at \(x=-2\), and it is negative on \((0,1)\). Therefore, \(f\) is increasing on \((-3,0)\) and decreasing on \((0,1)\). 2. At \(x=-2\), \(f'\) is zero but does not change sign, so the horizontal tangent is not an extremum. Because \(f'\) has a local minimum there, the concavity of \(f\) changes, making \(x=-2\) a stationary-inflection input. At \(x=0\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 3. The graph of \(f'\) has local extrema at \(x=-2\) and approximately \(x=-0.7\). Therefore, \(f\) has inflection points at those x-values. 4. Since \(f\) increases up to \(x=0\) and decreases afterward, its greatest value on the displayed interval occurs at \(x=0\).

a) Increasing on \((-3,0)\); decreasing on \((0,1)\). b) \(x=-2\): stationary inflection point; \(x=0\): local maximum. c) Inflection-point x-coordinates: \(-2\) and approximately \(-0.7\). d) \(x=0\).

- Find the zeros of \(f^{\prime}\). - Use the sign of \(f^{\prime}\) on each interval to determine whether \(f\) increases or decreases. - A sign change at a zero indicates a local extremum. - Turning points of \(f^{\prime}\) indicate changes in the concavity of \(f\).

1. The zeros of \(f^{\prime}\) are \(x=-3\), \(x=-1\), and \(x=2\). These are the x-values where \(f\) has horizontal tangents. The graph of \(f^{\prime}\) has a local maximum at approximately \(x=-2.1\) and a local minimum at approximately \(x=0.8\), so a corresponding graph of \(f\) changes concavity at those x-values. Any vertical shift with these slope and concavity properties is possible. 2. At \(x=-3\), \(f^{\prime}\) changes from negative to positive, so \(f\) has a local minimum. At \(x=-1\), it changes from positive to negative, so \(f\) has a local maximum. At \(x=2\), it changes from negative to positive, so \(f\) has a local minimum. There are no stationary inflection points because the derivative changes sign at every zero.

1. One possible graph has local minima at \(x=-3\) and \(x=2\), a local maximum at \(x=-1\), and inflection points at approximately \(x=-2.1\) and \(x=0.8\). 2. Horizontal tangents occur at \(x=-3\) (local minimum), \(x=-1\) (local maximum), and \(x=2\) (local minimum). There are no stationary inflection points.

- Use the inflection point on the y-axis to find the constant term. - At a marked local extremum, use both its coordinates and the condition that the derivative is \(0\).

1. The graph's inflection point is \((0, 1)\), so \(c=1\). 2. The graph has a local maximum at \((-1, 3)\) and a local minimum at \((1, -1)\). Since \(f'(x)=3ax^2+b\), the zero slope at \(x=1\) gives \(3a+b=0\), so \(b=-3a\). 3. Using \(f(1)=-1\), \(a-3a+1=-1\), so \(a=1\) and \(b=-3\). 4. Therefore, \(f(x)=x^3-3x+1\).

\(f(x)=x^3-3x+1\)

- At an inflection point of \(f\), the slope represented by \(f'\) changes from increasing to decreasing or vice versa. - Where does the graph of \(f'\) have local maxima or minima? - How does the monotonicity of \(f'\) determine the concavity of \(f\)? - Concave-down behavior occurs where \(f'\) is decreasing.

1. Inflection points of \(f\) occur where \(f'\) has local extrema, because the monotonicity of \(f'\) changes there. 2. The graph of \(f'\) has local extrema at \(x=-2\), \(x=0\), and \(x=2\). Therefore, these are the x-coordinates of the inflection points of \(f\). 3. The displayed x-range is \([-3,4]\). The graph of \(f'\) is decreasing on \([-3,-2)\) and \((0,2)\), so \(f\) is concave down on those intervals.

a) \(x=-2\), \(x=0\), and \(x=2\) b) \([-3, -2)\) and \((0, 2)\)

- Factor first to get the intercepts efficiently. - Use the zeros and signs of \(f'\) to organize the monotonicity pieces. - Use the sign of \(f''\) to organize the concavity pieces and locate the inflection point. - The leading cubic term determines the two ends of the blueprint.

1. Factoring gives \(f(x)=\frac12x(x^2-12)\). The x-intercepts are \((0,0)\) and \((\pm2\sqrt3,0)\). 2. \(f'(x)=\frac32x^2-6\), so the critical numbers are \(x=\pm2\). Since \(f''(x)=3x\), the second derivative test gives a local maximum at \((-2,8)\) and a local minimum at \((2,-8)\). 3. The second derivative changes sign at \(x=0\), so \((0,0)\) is the inflection point. 4. The graph increases on \(( -\infty,-2)\), decreases on \((-2,2)\), and increases on \((2,\infty)\). It is concave down on \(( -\infty,0)\) and concave up on \((0,\infty)\). 5. Because the leading term is positive cubic, the left end goes to \(-\infty\) and the right end goes to \(\infty\). These facts determine the requested blueprint.

a) Intercepts: \((0,0)\), \((\pm2\sqrt3,0)\). Local maximum: \((-2,8)\). Local minimum: \((2,-8)\). Inflection point: \((0,0)\). b) Increasing on \(( -\infty,-2)\), decreasing on \((-2,2)\), increasing on \((2,\infty)\); concave down for \(x<0\), concave up for \(x>0\). Left-to-right: left end down → maximum \((-2,8)\) → inflection \((0,0)\) → minimum \((2,-8)\) → right end up.

- Multiplicity at a zero can help predict whether the graph crosses with a horizontal tangent. - A critical number where \(f'\) does not change sign is not a local extremum. - Split the real line at the zeros of \(f''\) to get the concavity intervals. - Assemble the blueprint in increasing x-order rather than treating the features as unrelated facts.

1. Factoring gives \(f(x)=x^3(x-4)\). Thus \(x=0\) is a zero of multiplicity \(3\), and \(x=4\) is a simple zero. Since the leading term is \(x^4\), \(f(x)\to\infty\) as \(x\to\pm\infty\). 2. \(f'(x)=4x^2(x-3)\). The derivative is negative for \(x<3\) except for the non-sign-changing zero at \(x=0\), and positive for \(x>3\). Therefore, \((3,-27)\) is the only local extremum, a local minimum. 3. \(f''(x)=12x(x-2)\). It is positive for \(x<0\), negative for \(0<x<2\), and positive for \(x>2\). Thus the inflection points are \((0,0)\) and \((2,-16)\). Because \(f'(0)=0\), \((0,0)\) is a stationary inflection point. 4. The graph decreases and is concave up until \((0,0)\), continues decreasing and becomes concave down until \((2,-16)\), continues decreasing and becomes concave up until the minimum \((3,-27)\), then increases concave up and crosses again at \((4,0)\).

a) Zeros: \(x=0\) (multiplicity \(3\)) and \(x=4\); both ends go to \(\infty\). b) Only local extremum: local minimum \((3,-27)\). c) Inflection points \((0,0)\) and \((2,-16)\); \((0,0)\) is stationary. d) Decreasing on \(( -\infty,3)\), increasing on \((3,\infty)\); concave up on \(( -\infty,0)\) and \((2,\infty)\), concave down on \((0,2)\). Significant order: stationary inflection \((0,0)\), inflection \((2,-16)\), minimum \((3,-27)\), zero \((4,0)\).

- Match local extrema of one graph with zeros of the other. - Compare increasing and decreasing intervals with the sign of the possible derivative. - Use the polynomial degrees as a check.

1. The blue graph \(k\) has a local maximum at \(x = -1\) and a local minimum at \(x = 3\). 2. The red graph \(m\) is zero at \(x = -1\) and \(x = 3\). It is negative on \((-1, 3)\), where \(k\) decreases, and positive outside that interval, where \(k\) increases. 3. Therefore, \(k\) represents \(f\), and \(m\) represents \(f'\).

The blue graph \(k\) represents \(f\), and the red graph \(m\) represents \(f'\). The zeros and signs of \(m\) match the local extrema and monotonicity of \(k\).

- Match horizontal tangents with zeros of the derivative. - Compare increasing and decreasing intervals with the sign of the derivative. - Use symmetry and the locations of greatest slopes as additional checks.

1. The function \(f\) is a parabola with vertex at \(x = 0\). Its derivative must be a line through the origin, so \(f\) matches \(r\). 2. The function \(g\) is cubic with a local minimum at \(x = -2\) and a local maximum at \(x = 2\). Its derivative must be zero at those values and positive between them, so \(g\) matches \(p\). 3. The function \(h\) is sinusoidal. Its derivative is a cosine graph with the same period and is largest at \(x = 0\), where \(h\) has its greatest positive slope. Thus, \(h\) matches \(q\).

\(f \rightarrow r\); \(g \rightarrow p\); \(h \rightarrow q\).

- Use the sign of \(f'\) to determine increasing intervals. - Use sign changes at zeros of \(f'\) to classify local extrema. - A local extremum of \(f'\) indicates a possible inflection point of \(f\).

1. On the displayed interval, the derivative is positive for \(-1 < x < 1\) and \(3 < x < 5\), so \(f\) is strictly increasing on \([-1, 1]\) and \([3, 5]\). 2. At \(x = 1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. At \(x = 3\), it changes from negative to positive, so \(f\) has a local minimum. 3. The derivative graph has a local minimum at \(x = 2\). Thus, \(f''(2) = 0\) and \(f''\) changes from negative to positive. Therefore, \(f\) changes from concave down to concave up and has an inflection point at \(x = 2\).

a) \([-1, 1]\) and \([3, 5]\) b) Local maximum at \(x = 1\); local minimum at \(x = 3\) c) \(f\) has an inflection point at \(x = 2\), changing from concave down to concave up.

- Locate the local maximum and local minimum of \(g\). - The derivative must be zero at those inputs. - Determine the sign of the derivative between the extrema. - Select the candidate with the matching sign pattern.

1. The graph of \(g\) has a local maximum near \(x=-1.73\) and a local minimum near \(x=1.73\). 2. Therefore, \(g^{\prime}\) must have zeros near those inputs. Both candidate graphs have the required zeros. 3. Between the two extrema, \(g\) is decreasing, so \(g^{\prime}(x)<0\) there. 4. Graph 3 lies below the x-axis between its zeros, while graph 4 lies above it. Therefore, graph 3 represents \(g^{\prime}\).

Graph 3 represents \(g^{\prime}\).

- Mark every input where the graph has a horizontal tangent line. - Determine whether \(f\) increases or decreases on each interval. - The derivative of a fourth-degree polynomial is cubic. - A steeper graph corresponds to a derivative value farther from zero.

1. Zeros of \(f^{\prime}\) occur where \(f\) has horizontal tangent lines. The graph has local minima at \(x=-2\) and \(x=2\) and a local maximum at \(x=0\). Thus, the zeros are \(x=-2\), \(x=0\), and \(x=2\). 2. For \(x<-2\), \(f\) is decreasing, so \(f^{\prime}(x)<0\). For \(-2<x<0\), \(f\) is increasing, so \(f^{\prime}(x)>0\). For \(0<x<2\), \(f\) is decreasing, so \(f^{\prime}(x)<0\). For \(x>2\), \(f\) is increasing, so \(f^{\prime}(x)>0\). 3. Since \(f\) is a fourth-degree polynomial, \(f^{\prime}\) is cubic. Its graph crosses the x-axis at \(-2\), \(0\), and \(2\) with the sign pattern negative, positive, negative, positive.

a) \(x=-2\), \(x=0\), and \(x=2\) b) The derivative graph is cubic. It lies below the x-axis where \(f\) is decreasing and above the x-axis where \(f\) is increasing.

- Use the given formula to justify behavior beyond the plotted window. - Determine the derivative's sign from its numerator and denominator. - Compare how quickly numerator and denominator grow in magnitude for large inputs.

1. In the displayed interval, the plotted curve rises to the left of \(0\), falls to the right of \(0\), and has a horizontal tangent at \(x=0\). 2. The quotient rule gives \(f'(x)=\frac{-4x}{(1+0.5x^2)^2}\). 3. The derivative is zero only at \(x=0\). Its denominator is always positive, so \(f'(x)>0\) for \(x<0\) and \(f'(x)<0\) for \(x>0\). 4. For large \(|x|\), the denominator grows on the order of \(x^4\) while the numerator grows on the order of \(x\). Thus, \(f'(x)\to0\) as \(x\to\pm\infty\). 5. The derivative's signs and zero agree with the rise, fall, and flat point read from the graph.

In the displayed graph, \(f\) rises to the left of \(0\), falls to the right, and is flat at \(0\). The derivative formula confirms for all real \(x\) that \(f'>0\) for \(x<0\), \(f'<0\) for \(x>0\), and \(f'(0)=0\). Also, \(f'(x)\to0\) as \(x\to\pm\infty\).

- Match local extrema of one graph with zeros of the other. - Compare monotonicity with the sign of the possible derivative. - Read the tangent slope at \(x = 0\) and compare it with the other graph's value there.

1. The graph of \(f\) has a local maximum at \(x = -2\) and a local minimum at \(x = 2\). The graph of \(g\) is zero at those same values. Also, \(g\) is negative where \(f\) decreases and positive where \(f\) increases. Therefore, \(g = f'\). 2. The reverse assignment is impossible because \(g\) is quadratic, so its derivative would be linear, while \(f\) is cubic. 3. At \(x = 0\), the slope of \(f\) is approximately \(-1.2\). The graph of \(g\) gives \(g(0) \approx -1.2\), confirming the derivative relationship.

a) \(g = f'\). Its zeros match the local extrema of \(f\), and its sign matches where \(f\) increases or decreases. b) The slope of \(f\) at \(x = 0\) is approximately \(-1.2\), and \(g(0) \approx -1.2\).

- Differentiating the given formula lets you check an entire interval, not just a finite sketch. - Compare the signs of the numerator and denominator of \(f'\). - When comparing negative slopes, check their order on the number line.

1. The quotient rule gives \(f'(x)=\frac{2(1-x^2)}{(x^2+1)^2}\). 2. Statement (1) is true. The graph falls outward from its two turning points, and the derivative numerator is nonpositive when \(|x|\ge1\), including the stated displayed interval. 3. Statement (2) is false: \(f'(0)=2\), not \(1\). 4. Statement (3) is true: \(f'(2)=-\frac{6}{25}\) and \(f'(3)=-\frac{4}{25}\), so the first is smaller. 5. Statement (4) is true: the graph has a smooth maximum at \(1\), where \(f'(1)=0\).

(1) True; the graph decreases on the displayed outer intervals. (2) False; \(f'(0)=2\). (3) True; \(f'(2)=-\frac{6}{25}<-\frac{4}{25}=f'(3)\). (4) True; the graph has a smooth maximum at \(x=1\).

- Find the local extrema of \(f\). - The derivative must be zero at those inputs. - Use the increasing and decreasing intervals to determine the sign of the derivative. - The derivative of a cubic polynomial is quadratic.

1. The graph has a local minimum at \(x=-1\) and a local maximum at \(x=3\). Therefore, \(f^{\prime}\) must have zeros at \(x=-1\) and \(x=3\). 2. Options A and B have those two zeros. Option C is linear and has only one zero, and option D has different zeros. 3. Between \(-1\) and \(3\), the graph of \(f\) is increasing, so \(f^{\prime}(x)>0\) there. 4. At \(x=0\), option A gives \(0.9>0\), while option B gives \(-0.9<0\). Therefore, option A has the correct sign pattern.

A) \(f^{\prime}(x)=-0.3x^2+0.6x+0.9\)

- The derivative sign indicates whether the original graph is increasing or decreasing. - Horizontal tangent lines give zeros of the derivative. - Check the entire stated interval, including its endpoints. - Count all local extrema of the original graph.

1. Statement (1) is true. The graph is increasing at \(x=-1\), so \(f^{\prime}(-1)>0\). 2. Statement (2) is true. The graph has a local maximum at \(x=0\), so \(f^{\prime}(0)=0\). 3. Statement (3) is false. At \(x=0\), the derivative equals zero, and for \(0<x\le1.5\), the graph of \(f\) is decreasing, so \(f^{\prime}(x)<0\). 4. Statement (4) is true. The graph has a local minimum near \(x=1.73\), so \(f^{\prime}\) has a zero there. 5. Statement (5) is false. The graph is increasing at \(x=2\), so \(f^{\prime}(2)>0\). 6. Statement (6) is true. The graph has three horizontal tangent lines, so \(f^{\prime}\) has three zeros.

(1) True (2) True (3) False (4) True (5) False (6) True

- Local extrema of \(g\) give zeros of \(g^{\prime}\). - The derivative is negative where the function is decreasing. - Compare the signs of the tangent slopes at \(x=-1\) and \(x=1\).

1. The derivative is zero where the graph has horizontal tangent lines. The graph has a local maximum at \(x=0\) and a local minimum at \(x=2\). Thus, \(g^{\prime}(0)=0\) and \(g^{\prime}(2)=0\). 2. The graph of \(g\) is decreasing on \((0,2)\), so \(g^{\prime}(x)<0\) throughout that interval. 3. At \(x=-1\), the graph is increasing, so \(g^{\prime}(-1)>0\). At \(x=1\), it is decreasing, so \(g^{\prime}(1)<0\). Therefore, \(g^{\prime}(-1)>g^{\prime}(1)\).

a) \(x=0\) and \(x=2\) b) Negative: \(g^{\prime}(x)<0\) on \((0,2)\). c) \(g^{\prime}(-1)>g^{\prime}(1)\).

- Look at the tangent direction at each specified input. - Compare the graph's height as you move from left to right. - Check whether the tangent is horizontal at any other points in the displayed interval.

1. Statement (1) is true. The graph has a local maximum at \(x=-2\), so its tangent is horizontal. 2. Statement (2) is false. The graph is decreasing at \(x=0\), so the tangent slope there is negative. 3. Statement (3) is true. The graph is decreasing throughout \([-1,1]\), and its tangents have negative slopes there. 4. Statement (4) is true. The displayed graph has horizontal tangents at \(x=-2\) and \(x=2\), and nowhere else in the displayed interval.

(1) True; the tangent is horizontal at the high point \(x=-2\). (2) False; \(h\) falls at \(x=0\), so \(h'(0)<0\). (3) True; the curve falls on \([-1,1]\), so its tangent slopes are negative there. (4) True; the horizontal tangents in the displayed interval occur at \(x=-2\) and \(x=2\).

- At a smooth local maximum or minimum, the derivative is zero. - Compare where each function increases or decreases with where each derivative is positive or negative. - A quadratic function has a linear derivative, while a cubic function has a quadratic derivative.

1. Graph (1) is a downward-opening parabola with a local maximum at \(x=2\). Its derivative must be positive before \(x=2\), zero at \(x=2\), and negative after \(x=2\). Graph (B), the decreasing line \(y=-x+2\), has this sign pattern. Therefore, (1) matches (B). 2. Graph (2) is a cubic with a local maximum at \(x=-2\) and a local minimum at \(x=2\). Its derivative must be zero at \(x=-2\) and \(x=2\), positive outside those values, and negative between them. Graph (A), the upward-opening parabola \(y=0.75x^2-3\), has this sign pattern. Therefore, (2) matches (A).

a) (1) matches (B); (2) matches (A). b) The critical point of (1) at \(x=2\) matches the zero and sign change of (B). The critical points of (2) at \(x=-2\) and \(x=2\) match the zeros and sign changes of (A).

- Count all horizontal tangent lines on the graph of \(g\). - Match their x-coordinates to zeros of the candidates. - Check whether \(g\) is increasing or decreasing on each interval. - Eliminate any candidate with the wrong sign.

1. The fourth-degree graph of \(g\) has local minima at \(x=-2\) and \(x=2\) and a local maximum at \(x=0\). Therefore, \(g^{\prime}\) must have zeros at \(x=-2\), \(x=0\), and \(x=2\). 2. Graph A has only two zeros, so it cannot represent \(g^{\prime}\). 3. Graph C has three zeros, but they occur at different inputs. Its sign pattern also does not match: for \(x<-2\), the graph of \(g\) is decreasing, so \(g^{\prime}(x)\) must be negative. 4. Graph B has the required zeros and sign pattern. Therefore, graph B represents \(g^{\prime}\).

Graph B represents \(g^{\prime}\). Graph A has too few zeros, and graph C has zeros at the wrong inputs and the wrong sign pattern.

- Zeros of \(f'\) indicate points where \(f\) has a horizontal tangent. - A sign change in \(f'\) determines whether a local maximum or minimum occurs. - Consider what happens when \(f'\) equals zero but keeps the same sign on both sides. - Make sure you are interpreting the graph of the derivative, not the graph of \(f\).

1. Local extrema occur at zeros of \(f'\) where the sign changes. At \(x=-3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x=1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 2. At \(x=4\), \(f'(4)=0\), but \(f'\) is negative on both sides of \(x=4\). Thus, \(f\) continues decreasing and does not have a local extremum. Because \(f'\) changes from increasing to decreasing at \(x=4\), the concavity of \(f\) changes there. Therefore, \(f\) has a stationary inflection point at \(x=4\).

a) Local minimum at \(x=-3\); local maximum at \(x=1\). b) A stationary inflection point occurs at \(x=4\).

- A zero of the derivative alone does not guarantee a local extremum. - Check whether the graph of \(f'\) crosses or only touches the \(x\)-axis. - A stationary inflection point has a horizontal tangent but is not a local extremum.

1. Although \(f'(2)=0\), the derivative does not change sign at \(x=2\). Therefore, \(f\) does not have a local extremum there. 2. At \(x=-1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum at \(x=-1\). 3. At \(x=2\), \(f'\) has a local maximum with value \(0\). Thus, \(f'(2)=0\) without a sign change, while \(f'\) changes from increasing to decreasing. Therefore, the concavity of \(f\) changes and \(f\) has a stationary inflection point at \(x=2\).

a) No. The derivative does not change sign at \(x=2\). b) Local maximum at \(x=-1\). c) A stationary inflection point occurs at \(x=2\).

- The slope of \(f'\) represents \(f''\). - The x-coordinates of inflection points of \(f\) occur where \(f'\) changes from increasing to decreasing or vice versa. - Where is the graph of \(f'\) increasing? - Where is the graph of \(f'\) decreasing?

1. The graph of \(f'\) has local extrema at \(x=-2\) and \(x=2\), so \(f\) has inflection points at those x-values. 2. On the displayed domain \([-4.5,4.5]\), \(f'\) is increasing on \((-2,2)\). Therefore, \(f\) is concave up on \((-2,2)\). 3. The graph of \(f'\) is decreasing on \([-4.5,-2)\) and \((2,4.5]\). Therefore, \(f\) is concave down on those intervals.

a) \(x=-2\) and \(x=2\) b) Concave up on \((-2,2)\); concave down on \([-4.5,-2)\) and \((2,4.5]\).

- Use each point’s position relative to the x-axis for the sign of \(f(x)\). - Use whether the graph is increasing, decreasing, or horizontal for the sign of \(f'(x)\). - Use concavity for the sign of \(f''(x)\). - At an inflection point, the concavity changes.

1. At \(A=(0, 0)\), the point lies on the x-axis, the graph is increasing, and it is concave down. Therefore, \(f(0)=0\), \(f'(0)>0\), and \(f''(0)<0\). 2. At \(B\), the graph has a local maximum above the x-axis and is concave down. Therefore, \(f(2)>0\), \(f'(2)=0\), and \(f''(2)<0\). 3. At \(C\), the graph is above the x-axis and decreasing. It changes concavity there. Therefore, \(f(4)>0\), \(f'(4)<0\), and \(f''(4)=0\).

Point \(A\): \(f(x)=0\), \(f'(x)>0\), \(f''(x)<0\) Point \(B\): \(f(x)>0\), \(f'(x)=0\), \(f''(x)<0\) Point \(C\): \(f(x)>0\), \(f'(x)<0\), \(f''(x)=0\)

- Which derivative determines the concavity of a function? - For a polynomial, a necessary condition must hold; a sufficient condition guarantees the conclusion. - How are local extrema of \(f'\) related to the slope of \(f\)? - Use the sign of \(f''\) to determine concavity.

1. Because \(f\) is a polynomial, a necessary condition for an inflection point is \(f''(x)=0\). The graph of \(f''\) has zeros at \(x=-0.5\), \(x=1.5\), and \(x=3.5\). At each zero, \(f''\) changes sign, which is sufficient for an inflection point. Therefore, all three values are x-coordinates of inflection points. 2. The graph of \(f\) is concave down where \(f''(x)<0\). In the displayed interval, this occurs on \([-1.5,-0.5)\) and \((1.5,3.5)\). 3. At \(x=1.5\), \(f''\) changes from positive to negative. Therefore, \(f'\) changes from increasing to decreasing and has a local maximum. Since \(f'\) gives the slope of \(f\), the slope of \(f\) is locally greatest there.

a) \(x=-0.5\), \(x=1.5\), and \(x=3.5\). For this polynomial, \(f''(x)=0\) is necessary, while a sign change in \(f''\) is sufficient. b) \([-1.5,-0.5)\) and \((1.5,3.5)\) c) At \(x=1.5\), \(f''\) changes from positive to negative, so \(f'\) has a local maximum and the slope of \(f\) is locally greatest.

- Compare the equation’s symmetry with the graph. - Evaluate \(f(0)\). - Find the critical numbers from the first derivative. - Compare the required extremum at \(x=0\) with the graph.

1. The function is even, so its graph must be symmetric about the y-axis. The displayed graph is shifted to the right and lacks this symmetry. 2. The y-intercept is \(f(0)=2\), but the displayed graph crosses the y-axis at \((0,-1)\). 3. The derivative is \(f'(x)=4x(x^2-2)\), so \(x=0\) is a critical number. Since \(f''(0)=-8<0\), the graph of \(f\) must have a local maximum at \((0,2)\). The displayed graph has no extremum at \(x=0\).

The graph is incorrect because it is not symmetric about the y-axis, its y-intercept is \(-1\) instead of \(2\), and it has no local maximum at \((0, 2)\).

- Determine the degree of \(f'\). - Compare the symmetry of \(f\), \(f'\), and the displayed graph. - Compare the derivative value at \(x=0\) with the graph. - Consider how the end behavior of a cubic differs from that of a parabola.

1. Differentiate: \(f'(x)=4x^3-4x\). 2. Degree: The derivative of a fourth-degree polynomial is a third-degree polynomial, but \(g\) is a parabola, so it is quadratic. 3. Symmetry: The function \(f\) is even, so \(f'\) must be odd and symmetric about the origin. The graph of \(g\) is symmetric about the y-axis. 4. Value at the origin: Since \(f'(0)=0\), the graph of \(f'\) must pass through the origin. The graph shown has \(g(0)=-2\).

Three valid reasons are: \(f'\) must be cubic, but \(g\) is quadratic; \(f'\) must be odd, but \(g\) is even; and \(f'(0)=0\), but \(g(0)=-2\).

- Determine the degree of \(f'\). - Factor \(f'\) and examine the multiplicity of its zero at \(x=0\). - Compare the signs of \(f'\) and \(g\) for negative \(x\)-values. - Relate the zeros of a derivative to the horizontal tangents of the original function.

1. Differentiate and factor: \(f'(x)=4x^3-12x^2=4x^2(x-3)\). 2. Degree: The derivative \(f'\) is cubic, but the graph of \(g\) is a parabola, so \(g\) is quadratic. 3. Behavior at \(x=0\): The derivative \(f'\) has a double zero at \(x=0\) and does not change sign there. The graph of \(g\) crosses the \(x\)-axis at \(x=0\), changing from positive to negative. 4. Sign for negative inputs: For \(x<0\), \(4x^2>0\) and \(x-3<0\), so \(f'(x)<0\). The displayed graph has \(g(x)>0\) for \(x<0\).

Three valid reasons are: \(f'\) must be cubic, but \(g\) is quadratic; \(f'\) has a double zero and no sign change at \(x=0\), but \(g\) changes sign there; and \(f'(x)<0\) for every \(x<0\), but the graph shows \(g(x)>0\) for \(x<0\).

- When \(f'\) is increasing, \(f\) is concave up; when \(f'\) is decreasing, \(f\) is concave down. - The x-coordinates of inflection points of \(f\) occur where \(f'\) changes from increasing to decreasing or vice versa. - Restrict your conclusions to the displayed domain.

1. True. The graph of \(f'\) is increasing on \((-1,3)\), so \(f''(x)>0\) there. Therefore, \(f\) is concave up on that interval. 2. False. An inflection point of \(f\) occurs where \(f'\) changes from increasing to decreasing or from decreasing to increasing. The graph of \(f'\) has local extrema at \(x=-1\) and \(x=3\), not at \(x=0\). 3. True. The graph of \(f'\) is decreasing for \(3<x\le5.5\), so \(f''(x)<0\) there. Therefore, \(f\) is concave down.

1. True; \(f'\) is increasing on \((-1,3)\). 2. False; \(f'\) does not have a local extremum at \(x=0\). 3. True; \(f'\) is decreasing for \(3<x\le5.5\).

- A local extremum of \(f\) corresponds to a zero of \(f^{\prime}\). - At a local minimum, the derivative changes from negative to positive. - At a local maximum, the derivative changes from positive to negative. - Check both the number of zeros and the direction of each sign change.

1. At the local minimum \(x=-2\), the derivative must be zero and change sign from negative to positive. 2. At the local maximum \(x=2\), the derivative must be zero and change sign from positive to negative. 3. Graph C has only one zero, so it cannot represent the derivative. 4. Graph B has zeros at \(-2\) and \(2\), but its sign changes correspond to a maximum at \(-2\) and a minimum at \(2\), the reverse of the given information. 5. Graph A has zeros at \(-2\) and \(2\), with a negative-to-positive sign change at \(-2\) and a positive-to-negative sign change at \(2\). Therefore, graph A represents \(f^{\prime}\).

Graph A represents \(f^{\prime}\).

- Use the zeros to write a factored cubic with an unknown leading factor. - Substitute the additional point to find the factor. - Use the second and third derivatives for the inflection point. - Test \(f(-x)\) for origin symmetry. - A shift right by \(2\) replaces \(x\) with \(x-2\), and a shift up by \(1\) adds \(1\).

1. The zeros give \(f(x)=a x(x+3)(x-3)=a(x^3-9x)\). Using \(P=(1, -2)\), \(-2=-8a\), so \(a=\frac{1}{4}\). Thus, \(f(x)=\frac{1}{4}(x^3-9x)\). 2. The derivatives are \(f'(x)=\frac{3}{4}x^2-\frac{9}{4}\), \(f''(x)=\frac{3}{2}x\), and \(f'''(x)=\frac{3}{2}\). Since \(f''(0)=0\) and \(f'''(0)\ne0\), \(W=(0, 0)\) is an inflection point. Its tangent slope is \(f'(0)=-\frac{9}{4}\), so the tangent line is \(y=-\frac{9}{4}x\). 3. Since \(f(-x)=-f(x)\), the graph is symmetric about the origin. Translating right \(2\) and up \(1\) gives \(h(x)=f(x-2)+1=\frac{1}{4}((x-2)^3-9(x-2))+1\), whose inflection point is \(W'=(2,1)\).

a) \(f(x)=\frac{1}{4}(x^3-9x)\) b) \(W=(0, 0)\); tangent line: \(y=-\frac{9}{4}x\) c) \(f(-x)=-f(x)\); \(h(x)=\frac{1}{4}((x-2)^3-9(x-2))+1\)

- Use zeros and sign changes of \(f'\) to locate and classify local extrema of \(f\). - An inflection point of \(f\) corresponds to a local extremum of \(f'\). - Use the sign of \(f'\) on \([-1, 0]\) to compare the two function values. - Check whether the derivative graph is above or below the \(x\)-axis.

1. The zeros of \(f'\) are \(x=-4\) and \(x=2\). At \(x=-4\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x=2\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 2. An inflection point of \(f\) occurs where \(f'\) changes from increasing to decreasing. The graph of \(f'\) has its vertex at \(x=-1\), so \(f\) has an inflection point at \(x=-1\). 3. On \([-1, 0]\), the graph of \(f'\) is above the \(x\)-axis, so \(f\) is strictly increasing. Since \(0>-1\), it follows that \(f(0)>f(-1)\).

a) Local minimum at \(x=-4\); local maximum at \(x=2\). b) The inflection point occurs at \(x=-1\). c) Since \(f'(x)>0\) on \([-1, 0]\), \(f\) is strictly increasing there, so \(f(0)>f(-1)\).

- Which special points on the derivative graph identify x-coordinates of inflection points of the original function? - What must the derivative graph do when the original function is concave down? - Is the derivative positive or negative between \(x=-2\) and \(x=-1\)? - Use the sign of \(f'\) to determine whether \(f\) is increasing or decreasing.

1. The x-coordinates of inflection points of \(f\) occur where \(f'\) has local extrema. From the graph, these occur at approximately \(x=-2.3\) and \(x=2.3\). 2. The graph of \(f\) is concave down where \(f'\) is decreasing. The derivative graph decreases between its local maximum and local minimum, so \(f\) is concave down on approximately \((-2.3,2.3)\). 3. On \([-2,-1]\), the graph of \(f'\) lies above the x-axis, so \(f'(x)>0\). Therefore, \(f\) is increasing on this interval, which gives \(f(-1)>f(-2)\).

a) Approximately \(x=-2.3\) and \(x=2.3\) b) Approximately \((-2.3,2.3)\), because \(f'\) is decreasing there. c) Since \(f'(x)>0\) on \([-2,-1]\), \(f\) is increasing and \(f(-1)>f(-2)\).

- Horizontal tangent lines on \(k\) give zeros of \(k^{\prime}\). - Determine where \(k\) is increasing or decreasing. - The derivative of a cubic polynomial is quadratic. - Check the sign of each candidate between the two zeros.

1. The graph of \(k\) has a local maximum at \(x=0\) and a local minimum at \(x=2\). Therefore, \(k^{\prime}\) must have zeros at \(x=0\) and \(x=2\). 2. Panel d) is a line with only one zero. Also, the derivative of a cubic function must be quadratic, so panel d) cannot be correct. 3. Panels b) and c) are parabolas with zeros at \(x=0\) and \(x=2\). 4. The graph of \(k\) is decreasing for \(0<x<2\), so \(k^{\prime}(x)<0\) on that interval. 5. Panel b) lies below the x-axis between the zeros, while panel c) lies above it. Therefore, panel b) represents \(k^{\prime}\).

Panel b) represents \(k^{\prime}\).

- Use the sign of \(f'\) for monotonicity and extrema. - A local extremum of the derivative need not be a local extremum of the original function. - Compare \(f(2)\) and \(f(-1)\) from whether \(f\) rises or falls between those x-values.

1. True. On \((-3,-1)\), \(f'>0\), so \(f\) is strictly increasing. 2. False. Near \(x\approx0.8\), the graph of \(f'\) has a local minimum but is not zero. Thus \(f'\ne0\), so \(f\) has no local extremum there. Instead, the change from decreasing to increasing in \(f'\) gives an inflection point of \(f\). 3. True. At \(x=-3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 4. False. Since \(f'(x)<0\) on \((-1,2)\), \(f\) is strictly decreasing there, so \(f(2)<f(-1)\).

(1) True. (2) False. (3) True. (4) False.

- Inflection points of \(G\) correspond to local extrema of \(G^{\prime}\). - The value \(g(0)\) is the slope of the tangent to \(G\) at \(x=0\). - Use the sign changes of \(g\) to classify the local extrema of \(G\).

1. Inflection points of \(G\) occur where \(g=G^{\prime}\) changes from increasing to decreasing or vice versa. The graph of \(g\) has local extrema at approximately \(x=-1.2\) and \(x=1.2\). Therefore, those are the approximate inflection-point x-values of \(G\). 2. The tangent slope is \(G^{\prime}(0)=g(0)=0\), and the point is \((0,1)\). Thus, the tangent line is \(y=1\). 3. The derivative \(g\) changes from negative to positive at \(x=-2\), so \(G\) has a local minimum there. It changes from positive to negative at \(x=0\), so \(G\) has a local maximum at \((0,1)\). It changes from negative to positive at \(x=2\), so \(G\) has another local minimum there. Since \(g\) is odd and \(G(0)=1\), the graph of \(G\) is symmetric about the y-axis.

a) Approximately \(x=-1.2\) and \(x=1.2\) b) \(y=1\) c) Local minima at \(x=-2\) and \(x=2\), and a local maximum at \((0,1)\); the graph is symmetric about the y-axis.

- Find where each function has a horizontal tangent. - Compare where each function increases or decreases with the sign of each possible derivative. - The derivative of a quadratic function is linear. - The derivative of a cubic function is quadratic.

1. The graph of \(f\) is an upward-opening parabola with vertex at \(x=2\). Therefore, \(f'\) must be zero at \(x=2\), negative for \(x<2\), and positive for \(x>2\). The linear graph (1) has this behavior, so \(f'\) is graph (1). 2. The graph of \(g\) has a local maximum to the left of \(x=0\) and a local minimum between \(x=3\) and \(x=4\). Its derivative must be zero at those two \(x\)-coordinates and negative between them. The upward-opening parabola (2) has this behavior, so \(g'\) is graph (2).

Graph (1) represents \(f'\), and graph (2) represents \(g'\).

- Which features of the displayed curve would help determine its equation? - How can you check a proposed derivative formula against more than one visible feature? - What different information about \(f\) comes from the height and from the steepness of its derivative graph?

1. The graph of \(f'\) has zeros at \(x=-1\) and \(x=3\), so write \(f'(x)=a(x+1)(x-3)\). The vertex is \((1,4)\), so \(4=a(2)(-2)\), giving \(a=-1\). Thus \(f'(x)=-x^2+2x+3\). 2. The derivative is negative on \((-2,-1)\), positive on \((-1,3)\), and negative on \((3,4)\). Therefore, \(f\) decreases on \((-2,-1)\) and \((3,4)\), and increases on \((-1,3)\). 3. At \(x=-1\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x=3\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 4. Since \(f''(x)=-2x+2\), \(f''\) changes from positive to negative at \(x=1\). Thus \(f\) is concave up for \(x<1\), concave down for \(x>1\), and has an inflection point at \(x=1\).

a) \(f'(x)=-x^2+2x+3\) b) Increasing on \((-1,3)\); decreasing on \((-2,-1)\) and \((3,4)\) c) Local minimum at \(x=-1\); local maximum at \(x=3\) d) Inflection point at \(x=1\); concavity changes from up to down.

- Read the sign of the derivative from its position relative to the x-axis. - Use sign changes at derivative zeros to classify extrema of the original function. - Use the three zeros to write the cubic derivative in factored form. - Inflection points of \(f\) occur where the derivative switches between increasing and decreasing.

1. The derivative is negative on \((-6,-4)\) and \((1,5)\), and positive on \((-4,1)\) and \((5,7)\). Thus \(f\) decreases on \((-6,-4)\) and \((1,5)\), and increases on \((-4,1)\) and \((5,7)\). 2. At \(x=-4\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x=1\), it changes from positive to negative, so \(f\) has a local maximum. At \(x=5\), it changes from negative to positive, so \(f\) has a local minimum. 3. The zeros are \(-4\), \(1\), and \(5\), so \(f'(x)=a(x+4)(x-1)(x-5)\). The graph shows \(f'(0)=2\), so \(2=20a\), giving \(a=0.1\). Hence \(f'(x)=0.1(x+4)(x-1)(x-5)\). 4. Inflection points of \(f\) occur where \(f'\) has local extrema. Differentiate: \(f''(x)=0.1(3x^2-4x-19)\). Solving \(f''(x)=0\) gives \(x=\frac{2-\sqrt{61}}{3}\) and \(x=\frac{2+\sqrt{61}}{3}\). The sign of \(f''\) changes at both values, so both are inflection-point x-coordinates.

a) Increasing on \((-4,1)\) and \((5,7)\); decreasing on \((-6,-4)\) and \((1,5)\) b) Local minima at \(x=-4\) and \(x=5\); local maximum at \(x=1\) c) \(f'(x)=0.1(x+4)(x-1)(x-5)\) d) \(x=\frac{2-\sqrt{61}}{3}\) and \(x=\frac{2+\sqrt{61}}{3}\)

- Use the sign of the derivative throughout the displayed interval. - A function that is strictly decreasing on a closed interval has its largest and smallest values at opposite endpoints. - Use the derivative graph's intercepts to determine its quadratic equation. - Concavity of \(f\) follows from whether \(f'\) is decreasing or increasing.

1. The graph of \(f'\) is below the x-axis for \(-2<x<2\), so \(f\) is strictly decreasing on \((-2,2)\). 2. Since \(f\) is continuous and strictly decreasing across the closed interval, its absolute maximum occurs at \(x=-2\) and its absolute minimum occurs at \(x=2\). 3. The zeros are \(x=-2\) and \(x=2\), so \(f'(x)=a(x-2)(x+2)\). The graph shows \(f'(0)=-2\), so \(-4a=-2\), giving \(a=\frac12\). Hence \(f'(x)=\frac12x^2-2\). 4. Since \(f''(x)=x\), \(f''<0\) on \((-2,0)\) and \(f''>0\) on \((0,2)\). Thus \(f\) is concave down on \((-2,0)\), concave up on \((0,2)\), and has an inflection point at \(x=0\).

a) \(f\) is strictly decreasing on \((-2,2)\). b) Absolute maximum at \(x=-2\); absolute minimum at \(x=2\). c) \(f'(x)=\frac12x^2-2\). d) Concave down on \((-2,0)\); concave up on \((0,2)\); inflection point at \(x=0\).

- Find the maximum of the given function. - Compare its exact maximum height with the marked graph maxima. - A multiplier outside the function creates a vertical scale.

1. The derivative of \(h\) is \(h'(x)=(2-x)e^{-0.5x}\). It changes from positive to negative at \(x=2\), so the maximum occurs there. Its value is \(h(2)=\frac{4}{e}\), which matches graph q. 2. All three graphs have the same zero and maximum x-coordinate, so they differ by a vertical scale factor. Their marked maximum heights are \(\frac{2}{e}\), \(\frac{4}{e}\), and \(\frac{6}{e}\). 3. Graph p has half the height of q, so its equation is \(xe^{-0.5x}\). Graph r has \(1.5\) times the height of q, so its equation is \(3xe^{-0.5x}\). 4. The family is \(f_k(x)=kxe^{-0.5x}\), where \(k>0\).

a) Graph q b) p: \(xe^{-0.5x}\); r: \(3xe^{-0.5x}\) c) \(f_k(x)=kxe^{-0.5x}\), \(k>0\)

- Use zeros and sign changes of \(f^{\prime}\) to classify local extrema of \(f\). - The function decreases where its derivative is negative. - Local extrema of \(f^{\prime}\) correspond to inflection points of \(f\). - Use amplitude and period to determine the cosine parameters.

1. The derivative is \(0\) at \(x=1.5\) and \(x=4.5\). It changes from positive to negative at \(x=1.5\), so \(f\) has a local maximum there. It changes from negative to positive at \(x=4.5\), so \(f\) has a local minimum there. 2. The function \(f\) is strictly decreasing where \(f^{\prime}<0\), which is \((1.5,4.5)\). 3. Inflection points of \(f\) occur where \(f^{\prime}\) changes from increasing to decreasing or vice versa. The periodic derivative has extrema at \(x=0\), \(x=3\), and \(x=6\), so these are the requested x-values. 4. The amplitude is \(2\), and the period is \(6\). Thus, \(a=2\) and \(b=\frac{2\pi}{6}=\frac{\pi}{3}\). Therefore, \(f^{\prime}(x)=2\cos(\frac{\pi}{3}x)\).

a) Local maximum at \(x=1.5\); local minimum at \(x=4.5\) b) \((1.5,4.5)\) c) \(x=0\), \(x=3\), and \(x=6\) d) \(f^{\prime}(x)=2\cos(\frac{\pi}{3}x)\)

- Find the horizontal tangent lines on the graph of \(f\). - Use increasing and decreasing intervals to determine the derivative sign. - Identify where \(f\) has its most negative slope. - The derivative of a cubic polynomial is quadratic.

1. The derivative is zero at the local extrema of \(f\). From the graph, these occur near \(x=-1.4\) and \(x=1.4\). 2. The graph of \(f\) is increasing for \(x<-1.4\), decreasing for \(-1.4<x<1.4\), and increasing for \(x>1.4\). Therefore, \(f^{\prime}\) is positive, then negative, then positive across those intervals. 3. Since \(f\) is cubic, \(f^{\prime}\) is an upward-opening parabola. Its minimum occurs at \(x=0\), where the graph of \(f\) has its most negative slope.

1) The zeros are approximately \(x=-1.4\) and \(x=1.4\). 2) The derivative graph is an upward-opening parabola, positive outside the two zeros and negative between them, with a minimum at \(x=0\).

- Use the sign of \(f'\) to determine whether \(f\) rises or falls. - A zero of \(f'\) gives an extremum only when the derivative changes sign. - For inflection points, track whether \(f'\) itself is increasing or decreasing.

1. Statement (1) is true. At \(x=0\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 2. Statement (2) is false. On \((0,3)\), \(f'>0\), so \(f\) is strictly increasing. 3. Statement (3) is false. An inflection point of \(f\) occurs where \(f'\) changes from increasing to decreasing or vice versa. At \(x=0\), \(f'\) is increasing through a zero, so the concavity of \(f\) does not change there. 4. Statement (4) is true. At both \(x=-3\) and \(x=3\), \(f'\) changes from positive to negative, so \(f\) has a local maximum.

(1) True (2) False (3) False (4) True

- Compare the y-intercept, sign, and end behavior of the formula with the four graphs. - Horizontal tangents of \(f\) correspond to zeros of \(f'\). - Think about how differentiating an even graph affects symmetry. - Use the derivative formula only after making the graphical prediction.

1. Since \(f(0)=2\), \(f(x)>0\), and \(f(x)\to0\) as \(x\to\pm\infty\), graph (1) represents \(f\). 2. Zeros of \(f'\) occur where \(f\) has horizontal tangents. Graph (1) has exactly one horizontal tangent, at its maximum \(x=0\). Thus \(f'\) has exactly one zero at \(x=0\). 3. The function \(f\) is even, so its derivative \(f'\) is odd. 4. Differentiating gives \(f'(x)=-\frac{8x}{(x^2+2)^2}\). Thus \(f'(x)>0\) for \(x<0\), \(f'(0)=0\), and \(f'(x)<0\) for \(x>0\). Also \(f'(x)\to0\) as \(x\to\pm\infty\), so its horizontal asymptote is \(y=0\).

a) Graph (1) b) One zero: \(x=0\) c) \(f\) is even; \(f'\) is odd. d) \(f'(x)=-\frac{8x}{(x^2+2)^2}\); \(f'>0\) for \(x<0\), \(f'<0\) for \(x>0\); horizontal asymptote \(y=0\).

- Classify an extremum from the sign change of the first derivative. - Concavity depends on whether the derivative is increasing or decreasing. - A limiting slope describes steepness, not the actual height of the graph.

1. At \(x=1\), \(f'\) changes from negative to positive, so \(f\) has a local minimum there. 2. Differentiate the given derivative: \(f''(x)=\frac{3-x}{2}e^{-x/2}\). Since the exponential factor is positive, \(f''>0\) for \(x<3\) and \(f''<0\) for \(x>3\). Thus \(f\) is concave up for \(x<3\), concave down for \(x>3\), and has an inflection point at \(x=3\). 3. As \(x\to\infty\), \(f'(x)\to0\), so the graph of \(f\) becomes nearly horizontal. A slope approaching \(0\) does not by itself determine the function values, so it does not by itself prove that \(f\) approaches a horizontal asymptote.

a) Local minimum at \(x=1\) b) Concave up for \(x<3\); concave down for \(x>3\); inflection point at \(x=3\) c) \(f'(x)\to0\), so the graph becomes nearly horizontal; this alone does not prove a horizontal asymptote.

- Compare zeros of a possible derivative with extrema of the original function. - The sign of a derivative must agree with where the original function increases or decreases. - Because \(f^{\prime\prime}=g^{\prime}\), determine where \(g\) rises, falls, and has horizontal tangents. - Changes in the concavity of \(g\) help locate extrema of \(g^{\prime}\).

1. The graph of \(f\) has a local maximum at \(x=0\), and \(g\) has a zero there with a sign change from positive to negative. Also, \(f\) increases where \(g>0\) and decreases where \(g<0\). Therefore, \(g=f^{\prime}\). 2. The graph of \(f\) is concave down where \(f^{\prime\prime}(x)<0\). Since \(f^{\prime\prime}=g^{\prime}\), this occurs where \(g\) is decreasing, approximately on \((-0.6, 0.6)\). 3. The graph of \(f^{\prime\prime}\) has zeros near \(x=-0.6\) and \(x=0.6\), where \(g\) has local extrema. It is negative between these zeros and positive outside them in the displayed domain. 4. The steepest downward slope of \(g\) occurs at \(x=0\), so \(f^{\prime\prime}\) has a local minimum there. The changes in concavity of \(g\) near \(x=-1\) and \(x=1\) produce local maxima of \(f^{\prime\prime}\) near those x-values.

a) \(g=f^{\prime}\) because the zero and sign of \(g\) agree with the local maximum and monotonicity of \(f\). b) Approximately \((-0.6, 0.6)\) c) The graph of \(f^{\prime\prime}\) is positive outside approximately \((-0.6, 0.6)\), negative inside, has zeros near \(x=\pm0.6\), a local minimum at \(x=0\), and local maxima near \(x=\pm1\).

- The sign of \(f^{\prime}\) determines monotonicity. - Sign changes of \(f^{\prime}\) classify local extrema. - The increasing or decreasing behavior of \(f^{\prime}\) determines concavity. - To compare the values at 0 and 4, look at the derivative sign between those inputs.

1. The derivative is positive on \((-4,-2)\) and \((4,6)\), so \(f\) is increasing there. It is negative on \((-2,4)\), so \(f\) is decreasing there. 2. At \(x=-2\), \(f^{\prime}\) changes from positive to negative, so \(f\) has a local maximum. At \(x=4\), it changes from negative to positive, so \(f\) has a local minimum. 3. The function \(f\) is concave up where \(f^{\prime}\) is increasing. The derivative has its minimum at \(x=1\) and increases for \(x>1\), so \(f\) is concave up on \((1,6)\) in the displayed interval. The inflection-point x-value is \(x=1\). 4. Since \(f^{\prime}(x)<0\) for \(0<x<4\), \(f\) is strictly decreasing on \([0,4]\). Therefore, \(f(4)<f(0)=4\).

a) Increasing on \((-4,-2)\) and \((4,6)\); decreasing on \((-2,4)\) b) Local maximum at \(x=-2\); local minimum at \(x=4\) c) Concave up on \((1,6)\); inflection point at \(x=1\) d) \(f(4)<4\)

- First use the zeros and sign of \(f'\) to organize the monotonicity and extrema. - The original function is concave up where its derivative graph is increasing and concave down where the derivative graph is decreasing. - The vertex of the derivative graph supplies the inflection input of the cubic. - Put the three significant x-values in order before writing the final blueprint.

1. The zeros of \(f'\) are \(x=-1\) and \(x=3\). The derivative is positive for \(x<-1\), negative for \(-1<x<3\), and positive for \(x>3\). Hence \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=3\). 2. The same sign chart gives increasing intervals \(( -\infty,-1)\) and \((3,\infty)\), with decreasing interval \((-1,3)\). 3. The graph of \(f'\) decreases until its vertex at \(x=1\) and increases afterward. Thus \(f''<0\) for \(x<1\) and \(f''>0\) for \(x>1\). Therefore, \(f\) is concave down on \(( -\infty,1)\), concave up on \((1,\infty)\), and has an inflection point at \(x=1\). 4. Left to right, the graph increases concave down to a local maximum at \(x=-1\), decreases concave down to the inflection input \(x=1\), continues decreasing concave up to a local minimum at \(x=3\), then increases concave up.

a) Local maximum at \(x=-1\); local minimum at \(x=3\). b) Increasing on \(( -\infty,-1)\) and \((3,\infty)\); decreasing on \((-1,3)\). c) Concave down on \(( -\infty,1)\); concave up on \((1,\infty)\); inflection-point x-coordinate \(1\). d) Left-to-right features: maximum at \(-1\), inflection at \(1\), minimum at \(3\), with the monotonicity/concavity sequence stated above.

- Consider the degree of the derivative of a quadratic function. - A local extremum of the original function occurs where its derivative is zero. - Read the derivative value at \(x=3\) to find the tangent slope. - The function and derivative have the same value where their graphs intersect.

1. Curve \(k\) is a parabola and curve \(m\) is a line. Since the derivative of a quadratic function is linear, \(k\) represents \(g\) and \(m\) represents \(g^{\prime}\). This also agrees with the local minimum of \(k\) and the zero of \(m\) at \(x=0\). 2. The slope of \(g\) at \(x=3\) is \(g^{\prime}(3)\). From curve \(m\), \(g^{\prime}(3)=1.5\). 3. The curves intersect for \(x>0\) at \(x=2\). At that point, \(g(2)=g^{\prime}(2)=1\).

a) Curve \(k\) represents \(g\), and curve \(m\) represents \(g^{\prime}\). b) \(1.5\) c) \(x=2\)

- Evaluate each function at \(x=0\) and find any zeros. - Differentiate each function and use the sign of the derivative to determine monotonicity and extrema. - Compare horizontal asymptotes and end behavior. - Check for symmetry about the y-axis.

1. For \(f_1(x)=e^x-2\), the y-intercept is \(-1\), the horizontal asymptote is \(y=-2\) as \(x\to-\infty\), and \(f_1'(x)=e^x>0\), so the function is strictly increasing. These features match graph a). 2. For \(f_2(x)=2e^{-x}\), the y-intercept is \(2\), \(f_2(x)\to0\) as \(x\to\infty\), and \(f_2'(x)=-2e^{-x}<0\), so the function is strictly decreasing. These features match graph b). 3. For \(f_3(x)=4e^{-0.2x^2}\), the function is even, and \(f_3'(x)=-1.6xe^{-0.2x^2}\). Thus, it increases for \(x<0\), decreases for \(x>0\), and has a maximum value of \(4\) at \(x=0\). These features match graph c). 4. For \(f_4(x)=(x-1)e^x\), the zero is \(x=1\), the y-intercept is \(-1\), and \(f_4'(x)=xe^x\). Thus, it decreases for \(x<0\), increases for \(x>0\), and has a local minimum at \(x=0\). These features match graph d).

a) \(f_1\); b) \(f_2\); c) \(f_3\); d) \(f_4\)

- Differentiate and factor before making the sign chart. - Use the derivative's sign to determine monotonicity. - The minimum of the derivative corresponds to the point where its slope changes from negative to positive. - Compare the requested derivative values with zero rather than relying only on the graph.

1. Differentiate: \(f'(x)=3x^2-12x+9=3(x-1)(x-3)\). Thus the critical numbers are \(x=1\) and \(x=3\). 2. The derivative is positive on \((-1,1)\) and \((3,5)\), and negative on \((1,3)\). Therefore, \(f\) increases on \((-1,1)\) and \((3,5)\), and decreases on \((1,3)\). 3. On the displayed domain, \(f'(x)\ge0\) for \(x\in[-1,1]\cup[3,5]\). 4. The quadratic \(f'\) has its vertex midway between its zeros, at \(x=2\). This is where the slope of \(f\) is smallest and where \(f''\) changes sign, so \(x=2\) is the inflection-point x-coordinate of \(f\). 5. \(f'(0)=9>0\), while \(f'(2)=-3<0\).

a) \(x=1\) and \(x=3\) b) Increasing on \((-1,1)\) and \((3,5)\); decreasing on \((1,3)\) c) \(f'(x)=3(x-1)(x-3)\); \(f'(x)\ge0\) on \([-1,1]\cup[3,5]\) d) \(x=2\); it is the inflection-point x-coordinate of \(f\). e) \(f'(0)>0\) and \(f'(2)<0\)

- A negative-to-positive sign change of the derivative gives a local minimum. - Inflection points of \(f\) correspond to local extrema of \(f'\). - Read increasing behavior of \(f\) from where \(f'\) is positive.

1. A local minimum of \(f\) occurs where \(f'\) changes from negative to positive. The graph shows this at \(x=-4\) and \(x=3\). 2. Inflection points of \(f\) occur where \(f'\) has local extrema. From the graph, these x-values are approximately \(x=-2.7\) and \(x=1.4\). 3. The statement is true. For every displayed \(x>3\), the graph of \(f'\) is above the x-axis, so \(f'(x)>0\) and \(f\) is strictly increasing there.

a) \(x=-4\) and \(x=3\) b) Approximately \(x=-2.7\) and \(x=1.4\) c) True; \(f'(x)>0\) for every displayed \(x>3\).

- Zeros of a derivative correspond to horizontal tangents of the function one level above it. - Compare derivative signs with increasing and decreasing intervals. - Build the derivative chain one link at a time.

1. Graph 3 has a local minimum at \(x=0\). Graph 1 is zero there and changes from negative to positive, so Graph 1 is the derivative of Graph 3. 2. Graph 1 has local extrema at \(x=-1\) and \(x=1\). Graph 2 is zero at those x-values, and its sign matches where Graph 1 increases or decreases. Therefore, Graph 2 is the derivative of Graph 1. 3. Thus Graph 3 is \(f\), Graph 1 is \(f'\), and Graph 2 is \(f''\).

Graph 1: \(f'\); Graph 2: \(f''\); Graph 3: \(f\)

- Symmetry about the y-axis restricts the powers that may appear. - An inflection point on the x-axis gives a function-value condition and a second-derivative condition. - Use the y-intercept to determine the constant term. - Solve the resulting two-equation system for the remaining coefficients.

1. Because \(f\) is even, write \(f(x)=ax^4+bx^2+c\). 2. The point \(M=(0, 4)\) gives \(c=4\). At the inflection point with \(x=1\), both \(f(1)=0\) and \(f''(1)=0\). The first condition gives \(a+b=-4\). Since \(f''(x)=12ax^2+2b\), the second gives \(12a+2b=0\), or \(6a+b=0\). Solving the system yields \(a=\frac{4}{5}\) and \(b=-\frac{24}{5}\). Therefore, \(f(x)=\frac{4}{5}x^4-\frac{24}{5}x^2+4\). 3. Since \(f''(0)=-\frac{48}{5}<0\), \(M=(0, 4)\) is a local maximum. Also, \(f'''(\pm1)=\pm\frac{96}{5}\ne0\), so \(x=-1\) and \(x=1\) are indeed inflection-point x-coordinates.

1. \(f(x)=ax^4+bx^2+c\) 2. \(f(x)=\frac{4}{5}x^4-\frac{24}{5}x^2+4\)

- Use the y-axis symmetry to reduce the number of coefficients. - A marked inflection point gives conditions on the function and its second derivative. - Find the critical numbers from the first derivative, then classify the extrema. - Keep \(\sqrt{3}\) as an exact value.

1. Because \(g\) is even, write \(g(x)=ax^4+bx^2+c\). The graph passes through the origin, so \(c=0\). The marked inflection point \(I_2=\left(1, \frac{5}{2}\right)\) gives \(g(1)=\frac{5}{2}\) and \(g''(1)=0\). Therefore, \(a+b=\frac{5}{2}\) and \(12a+2b=0\). Solving gives \(a=-\frac{1}{2}\) and \(b=3\), so \(g(x)=-\frac{1}{2}x^4+3x^2\). Since \(g'''(\pm1)=\mp12\ne0\), both marked points are indeed inflection points. 2. The derivative is \(g'(x)=-2x^3+6x=-2x(x^2-3)\). Thus, the critical numbers are \(x=0\) and \(x=\pm\sqrt{3}\). Since \(g''(0)=6>0\), \(x=0\) is a local minimum. At \(x=\pm\sqrt{3}\), \(g''(x)=-6x^2+6=-12<0\), so both are local maxima. Their common function value is \(g(\pm\sqrt{3})=-\frac{1}{2}(9)+3(3)=\frac{9}{2}\).

1. \(g(x)=-\frac{1}{2}x^4+3x^2\) 2. Local maxima: \((-\sqrt{3}, \frac{9}{2})\) and \((\sqrt{3}, \frac{9}{2})\)

- Use the y-axis symmetry to simplify the polynomial form. - Translate “meets the ground with a horizontal tangent” into a function-value condition and a derivative condition. - Use the center height to determine the constant term. - Solve the resulting system for the remaining coefficients.

1. Because the arch is symmetric about the y-axis, write \(f(x)=ax^4+bx^2+c\). 2. The center point \((0, 20)\) gives \(c=20\). 3. The endpoint \((20, 0)\) gives \(160{,}000a+400b+20=0\). 4. A horizontal tangent at \(x=20\) requires \(f'(20)=0\). Since \(f'(x)=4ax^3+2bx\), \(32{,}000a+40b=0\), so \(b=-800a\). 5. Substitution into the endpoint equation gives \(-160{,}000a=-20\), so \(a=\frac{1}{8000}\) and \(b=-\frac{1}{10}\). 6. Therefore, \(f(x)=\frac{1}{8000}x^4-\frac{1}{10}x^2+20\), for \(-20\le x\le20\). Also, \(f''(0)=-\frac{1}{5}<0\), confirming the maximum at the center.

\(f(x)=\frac{1}{8000}x^4-\frac{1}{10}x^2+20\), for \(-20\le x\le20\)

- Use the derivative's sign for one set of features and the derivative's rise or fall for another. - Treat zeros of the derivative and turning points of the derivative graph as different kinds of information. - Work through monotonicity and concavity separately before combining the conclusions.

1. The derivative is positive on \((-5,-3)\) and \((1,5)\), and negative on \((-3,1)\). Thus \(h\) increases on \((-5,-3)\) and \((1,5)\), and decreases on \((-3,1)\). 2. At \(x=-3\), \(h'\) changes from positive to negative, so \(h\) has a local maximum. At \(x=1\), it changes from negative to positive, so \(h\) has a local minimum. 3. The graph of \(h'\) decreases on \((-5,-1)\), increases on \((-1,3)\), and decreases on \((3,5)\). Therefore, \(h\) is concave down on \((-5,-1)\) and \((3,5)\), and concave up on \((-1,3)\). 4. Concavity changes at \(x=-1\) and \(x=3\), so these are inflection-point x-coordinates.

a) Increasing on \((-5,-3)\) and \((1,5)\); decreasing on \((-3,1)\). b) Local maximum at \(x=-3\); local minimum at \(x=1\). c) Concave down on \((-5,-1)\) and \((3,5)\); concave up on \((-1,3)\). d) Inflection-point x-coordinates: \(x=-1\) and \(x=3\).

- Use the sign of the derivative to determine how the original function moves. - At a zero of the derivative, compare the signs on the two sides before classifying it. - For concavity, look for where the derivative values trend downward or upward.

1. Because \(p'\) is positive on \((-4,-2)\), negative on \((-2,2)\), and positive on \((2,4)\), \(p\) increases on \((-4,-2)\) and \((2,4)\), and decreases on \((-2,2)\). 2. At \(x=-2\), \(p'\) changes from positive to negative, so \(p\) has a local maximum. At \(x=2\), it changes from negative to positive, so \(p\) has a local minimum. 3. The listed values of \(p'\) decrease from \(x=-4\) to \(x=0\), then increase from \(x=0\) to \(x=4\). Therefore, \(p\) is concave down on \((-4,0)\) and concave up on \((0,4)\).

a) Increasing on \((-4,-2)\) and \((2,4)\); decreasing on \((-2,2)\). b) Local maximum at \(x=-2\); local minimum at \(x=2\). c) Concave down on \((-4,0)\); concave up on \((0,4)\).

- A squared factor in \(f'\) gives a zero where the sign of \(f'\) may stay the same. - Differentiate the supplied expression for \(f'\) to get \(f''\); do not integrate. - Split the blueprint at every x-value where either monotonicity or concavity changes.

1. Since \(f'(x)=-(x+1)^2(x-2)\), the squared factor does not change sign at \(x=-1\). Thus \(f'>0\) on both sides of \(-1\). At \(x=2\), \(f'\) changes from positive to negative. 2. Therefore \((-1,0)\) is not a local extremum, while \((2,\tfrac{27}{4})\) is a local maximum. 3. Differentiate the given derivative: \(f''(x)=-3x^2+3=3(1-x^2)\). Hence \(f''=0\) at \(x=-1\) and \(x=1\), and its sign changes at both values. 4. At \(x=-1\), the horizontal tangent together with the concavity change makes \((-1,0)\) a stationary inflection point. At \(x=1\), \((1,4)\) is a nonstationary inflection point. 5. The graph is increasing/concave down on \(( -\infty,-1)\), increasing/concave up on \((-1,1)\), increasing/concave down on \((1,2)\), and decreasing/concave down on \((2,\infty)\).

Stationary inflection \((-1,0)\); nonstationary inflection \((1,4)\); local maximum \((2,\tfrac{27}{4})\). Blueprint: ↑/concave down on \(( -\infty,-1)\), ↑/concave up on \((-1,1)\), ↑/concave down on \((1,2)\), and ↓/concave down on \((2,\infty)\).

- Use all breakpoints where either derivative sign changes. - Endpoint values matter for absolute extrema even though endpoints are not interior critical numbers. - Keep local classification and absolute comparison as separate steps.

1. \((-4,-1)\): increasing and concave down. 2. At \((-1,5)\), \(f'\) changes positive to negative, so there is a local maximum. 3. \((-1,0)\): decreasing and concave down. 4. At \((0,3)\), concavity changes down to up, so this is an inflection point. 5. \((0,2)\): decreasing and concave up. 6. At \((2,1)\), \(f'\) changes negative to positive, so there is a local minimum. 7. \((2,4)\): increasing and concave up. 8. Compare the supplied candidate values at the endpoints and local extrema: \(0,5,1,4\). The absolute maximum is \(5\) at \(x=-1\); the absolute minimum is \(0\) at the endpoint \(x=-4\).

Blueprint: ↑/concave down on \((-4,-1)\); local max \((-1,5)\); ↓/concave down to inflection \((0,3)\); ↓/concave up to local min \((2,1)\); ↑/concave up on \((2,4)\). Absolute max \(5\) at \(x=-1\); absolute min \(0\) at \(x=-4\).

- Use the sign of the given second derivative to classify the critical point. - Remember that \(\sqrt{x^2}=|x|\) when evaluating limits at infinity. - Complete the square inside the radical to identify the intercepts of the slant asymptotes.

1. The first derivative is \(f^{\prime}(x)=\frac{4x+2}{\sqrt{4x^2+4x+10}}\). Setting the numerator equal to zero gives \(x=-\frac{1}{2}\). Since \(f(-\frac{1}{2})=3\) and \(f^{\prime\prime}(x)>0\), the graph has a local and absolute minimum at \((-\frac{1}{2}, 3)\). 2. The expression \(4x^2+4x+10\) is positive for every real \(x\), so \(f^{\prime\prime}(x)>0\) everywhere. The concavity never changes, so there are no inflection points. 3. Divide numerator and denominator of \(f^{\prime}\) by \(|x|\). For \(x\to\infty\), the limit is \(2\); for \(x\to-\infty\), the limit is \(-2\). 4. Complete the square: \(f(x)=\sqrt{(2x+1)^2+9}\). As \(x\to\infty\), \(f(x)-(2x+1)\to0\), so one asymptote is \(y=2x+1\). As \(x\to-\infty\), \(f(x)-(-2x-1)\to0\), so the other is \(y=-2x-1\).

1. Absolute minimum at \((-\frac{1}{2}, 3)\) 2. \(f^{\prime\prime}(x)>0\) for all \(x\), so there are no inflection points. 3. \(\lim_{x\to\infty}f^{\prime}(x)=2\) and \(\lim_{x\to-\infty}f^{\prime}(x)=-2\) 4. \(y=2x+1\) and \(y=-2x-1\)

- Compare the domains and look for vertical asymptotes. - Identify zeros and y-intercepts. - Differentiate each candidate to locate critical numbers and determine increasing or decreasing intervals. - Analyze end behavior in both directions and check symmetry.

1. For \(f_4(x)=(x^2-1)e^{-x}\), the zeros are \(x=-1\) and \(x=1\), and the y-intercept is \(-1\). Also, \(f_4'(x)=(-x^2+2x+1)e^{-x}\), so its critical numbers are \(x=1\pm\sqrt{2}\). Together with \(f_4(x)\to0\) as \(x\to\infty\), these features match graph \(G_1\). 2. For \(f_3(x)=\frac{e^x}{x}\), the domain excludes \(x=0\), giving a vertical asymptote there. Since \(f_3'(x)=\frac{e^x(x-1)}{x^2}\), the function decreases for \(x<1\) on each part of its domain and increases for \(x>1\). It approaches \(0\) from below as \(x\to-\infty\) and grows without bound as \(x\to\infty\). These features match graph \(G_2\). 3. For \(f_1(x)=2e^{-x^2}\), \(f_1'(x)=-4xe^{-x^2}\). Thus, the function increases for \(x<0\), decreases for \(x>0\), and has a maximum at \((0,2)\). It is even and approaches \(0\) as \(x\to\pm\infty\), matching graph \(G_3\). 4. For \(f_2(x)=x^2e^{-x}\), \(f_2'(x)=x(2-x)e^{-x}\). The function is nonnegative, has a double zero at \(x=0\), and has a local maximum at \(x=2\). None of the displayed graphs has these features, so \(f_2\) is unused.

\(G_1\): \(f_4(x)=(x^2-1)e^{-x}\); \(G_2\): \(f_3(x)=\frac{e^x}{x}\); \(G_3\): \(f_1(x)=2e^{-x^2}\); unused: \(f_2(x)=x^2e^{-x}\)

- Find each point where the graph of \(f\) has a horizontal tangent line. - Distinguish a local extremum from a horizontal inflection point. - Determine the sign of the derivative from the increasing and decreasing intervals. - A sign change produces a crossing; no sign change produces a touch.

1. The notable points give the following table: <table> <tr><th>Input \(x\)</th><th>Type of point on \(f\)</th><th>Value of \(f^{\prime}(x)\)</th></tr> <tr><td>\(0\)</td><td>Local minimum</td><td>\(0\)</td></tr> <tr><td>\(3\)</td><td>Horizontal inflection point</td><td>\(0\)</td></tr> </table> 2. For \(x<0\), the function is decreasing, so \(f^{\prime}(x)<0\). For \(0<x<3\), the function is increasing, so \(f^{\prime}(x)>0\). For \(x>3\), the function continues increasing, so \(f^{\prime}(x)>0\). 3. The derivative graph crosses the x-axis at \(x=0\), changing from negative to positive. At \(x=3\), it touches the x-axis from above without changing sign. One additional reference value is \(f^{\prime}(1)\approx0.8\).

The derivative has zeros at \(x=0\) and \(x=3\). At \(x=0\), it changes sign from negative to positive and crosses the x-axis. At \(x=3\), it remains positive on both sides and touches the x-axis.

- Zeros of a derivative correspond to horizontal tangents of the function one level above it. - Compare the sign of a possible derivative with where the preceding graph increases or decreases. - Apply the same derivative relationship repeatedly through the four panels.

1. Graph 4 has a local minimum at \(x=0\). Graph 3 is zero there and changes from negative to positive, so Graph 3 is the derivative of Graph 4. Thus Graph 4 is \(f\) and Graph 3 is \(f'\). 2. Graph 3 has local extrema near \(x=\pm1.6\). Graph 1 is zero at those x-values, and its sign matches where Graph 3 increases or decreases. Thus Graph 1 is \(f''\). 3. Graph 1 has local extrema at \(x=0\) and near \(x=\pm2.7\). Graph 2 is zero at those x-values, with signs matching the increase and decrease of Graph 1. Thus Graph 2 is \(f'''\).

Graph 1: \(f''\); Graph 2: \(f'''\); Graph 3: \(f'\); Graph 4: \(f\)

- What derivative behavior distinguishes a local minimum from a stationary inflection point? - What changes in the graph near an inflection point, and what does that mean for the derivative? - How much of a polynomial can its degree and zeros determine? - What additional information would fix a function's vertical position?

1. At the local minimum \(x=-\frac{3}{2}\), \(f'\) has a simple zero. At the stationary inflection point \(x=3\), \(f'\) has a double zero. Since \(f'\) is cubic, these are all its zeros. 2. The inflection-point x-coordinates of \(f\) are \(x=0\) and \(x=3\), so these are the zeros of \(f''\). Each is simple because the concavity changes at both values. 3. At \(x=0\), \(f''\) changes from positive to negative, so \(f'\) has a local maximum. At \(x=3\), \(f''\) changes from negative to positive, so \(f'\) has a local minimum. 4. The zeros and multiplicities give \(f'(x)=k\left(x+\frac{3}{2}\right)(x-3)^2\). The tangent slope at \(I\) gives \(f'(0)=2\), so \(2=k\left(\frac{3}{2}\right)(9)\), giving \(k=\frac{4}{27}\). Thus \(f'\) is uniquely determined. 5. If two polynomial functions have this same derivative, their difference has derivative zero everywhere. By the Mean Value Theorem, that difference is constant. Both must pass through \(I=(0,2)\), so the difference is zero. Therefore, \(f\) is uniquely determined without finding its formula.

a) Zeros of \(f'\): \(x=-\frac{3}{2}\), multiplicity \(1\); \(x=3\), multiplicity \(2\); zeros of \(f''\): \(x=0\) and \(x=3\), each with multiplicity \(1\). b) \(f'\) has a local maximum at \(x=0\) and a local minimum at \(x=3\). c) Yes. The zeros and \(f'(0)=2\) determine the derivative uniquely. Any two functions with that derivative differ by a constant (Mean Value Theorem), and \(f(0)=2\) fixes that constant.

- Compare the signs of the factored first derivative on the intervals separated by its zeros. - Differentiate the given derivative to check where the curve changes concavity. - The supplied function values place each turning or inflection point vertically; no formula for the original function is needed.

1. The zeros of \(f'\) are \(x=-2\) and \(x=1\). The factored derivative is negative on \((-\infty,-2)\), positive on \((-2,1)\), and negative on \((1,\infty)\). 2. Therefore, \((-2,0)\) is a local minimum and \((1,\frac{9}{2})\) is a local maximum. 3. Differentiate: \(f''(x)=-2x-1\). This is positive for \(x<-\frac{1}{2}\) and negative for \(x>-\frac{1}{2}\), so \((-\frac{1}{2},\frac{9}{4})\) is an inflection point. 4. From left to right, the graph decreases and is concave up before \(x=-2\); increases and is concave up between \(-2\) and \(-\frac{1}{2}\); increases and is concave down between \(-\frac{1}{2}\) and \(1\); then decreases and is concave down after \(x=1\).

Local minimum: \((-2,0)\). Inflection point: \((-\frac{1}{2},\frac{9}{4})\). Local maximum: \((1,\frac{9}{2})\). Decreasing and concave up on \((-\infty,-2)\); increasing and concave up on \((-2,-\frac{1}{2})\); increasing and concave down on \((-\frac{1}{2},1)\); decreasing and concave down on \((1,\infty)\).

- Use the ordered simple zeros of \(f'\) to build its sign chart. - Differentiate \(f'\) before building the concavity chart. - Merge the two ordered breakpoint lists only after both charts are correct. - The anchor and stated end behavior place the qualitative graph; no integration is needed.

1. The critical numbers are \(-2,-1,1,2\). Because all four zeros of \(f'\) are simple, its sign alternates. Since \(f'>0\) for large \(|x|\), the sign pattern is \(+,-,+,-,+\). 2. Therefore \(x=-2\) and \(x=1\) are local maxima, while \(x=-1\) and \(x=2\) are local minima. The maximal increasing intervals are \(( -\infty,-2)\), \((-1,1)\), and \((2,\infty)\); the maximal decreasing intervals are \((-2,-1)\) and \((1,2)\). 3. Differentiate: \(f''(x)=4x^3-10x=2x(2x^2-5)\). Thus the inflection inputs are \(-\sqrt{5/2},0,\sqrt{5/2}\). Each zero is simple, so concavity changes at each one. 4. Testing signs gives concave down on \(( -\infty,-\sqrt{5/2})\), concave up on \((-\sqrt{5/2},0)\), concave down on \((0,\sqrt{5/2})\), and concave up on \((\sqrt{5/2},\infty)\). 5. Left-to-right significant inputs are \(-2,-\sqrt{5/2},-1,0,1,\sqrt{5/2},2\). At \(x=0\), the graph passes through \((0,0)\) and changes concavity. Combine this order with the stated end behavior and interval directions.

Local maxima at \(x=-2,1\); local minima at \(x=-1,2\). Increasing on \(( -\infty,-2)\), \((-1,1)\), \((2,\infty)\); decreasing on \((-2,-1)\), \((1,2)\). Inflection inputs \(-\sqrt{5/2},0,\sqrt{5/2}\). Concave down on \(( -\infty,-\sqrt{5/2})\) and \((0,\sqrt{5/2})\); concave up on \((-\sqrt{5/2},0)\) and \((\sqrt{5/2},\infty)\). Significant order: \(-2,-\sqrt{5/2},-1,(0,0),1,\sqrt{5/2},2\), with left end down and right end up.