Free Math Worksheets

- 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 points, \(f_1\) 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 \([-2, 2]\).

- 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 points and monotonicity of \(f\), while the zero and signs of \(q\) match those of \(g\).

- 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 red graph 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 \(g'\).

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

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

- 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 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. \([-2, 2]\)

- 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

- 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).

- 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 \(h_2\). \(g\) matches \(h_1\).

- Zeros of \(f\) correspond to horizontal tangents of \(F\). - Where \(f>0\), \(F\) increases; where \(f<0\), \(F\) decreases. - Compare the graphs at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\).

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 antiderivative should. 2. Graph (1) is positive on \((0,\frac{\pi}{2})\), and graph (2) is increasing there. It 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 \(f\), and graph (2) is \(F\).

- Determine the sign of \(\ln(3)\). - A constant derivative produces a linear function. - The sign of the derivative determines whether the line rises or falls.

1. Since \(3>1\), \(\ln(3)>0\). Thus, \(f\) is a positive constant function. 2. Because \(F^{\prime}(x)=\ln(3)\) at every x-value, \(F\) must be a line with constant positive slope. 3. Only graph a has that form, so graph a could represent \(F\).

Graph a

- Zeros of \(\cos(x)\) correspond to horizontal tangents of \(F\). - Use the sign of \(\cos(x)\) to determine where \(F\) increases or decreases. - Recall which familiar function has derivative \(\cos(x)\).

1. Since \(F^{\prime}(x)=\cos(x)\), the slope of \(F\) is positive on \((0,\frac{\pi}{2})\), zero at \(x=\frac{\pi}{2}\), and negative on \((\frac{\pi}{2},\frac{3\pi}{2})\). 2. Thus, \(F\) must have a local maximum at \(x=\frac{\pi}{2}\) and a local minimum at \(x=\frac{3\pi}{2}\). At \(x=0\), its slope must be \(1\). 3. Graph a, \(F(x)=\sin(x)\), has all these properties. Therefore, graph a is a possible antiderivative.

Graph a

- An antiderivative of a linear function is quadratic. - Find the zero of \(f\). - Use the sign change of \(f\) to classify the corresponding extremum of \(F\).

1. The function \(f\) has a zero at \(x=2\). Since \(F^{\prime}=f\), an antiderivative must have a horizontal tangent there. 2. For \(x<2\), \(f<0\), so \(F\) decreases. For \(x>2\), \(f>0\), so \(F\) increases. Therefore, \(F\) has a local minimum at \(x=2\). 3. An antiderivative of a linear function is quadratic. Only graph a is an upward-opening parabola with vertex x-coordinate \(2\).

Graph a

- 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 values 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 values 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 \(x\to\pm\infty\).

- How are the critical points 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 blue graph \(f\) has a local maximum between \(x=-2\) and \(x=-1\), and a local minimum between \(x=1\) and \(x=2\). The green graph \(g\) is zero at those same \(x\)-coordinates. 2. The graph of \(f\) increases before the first critical point, decreases between the two critical points, and increases after the second critical point. 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 blue graph \(f\) represents the function, and the green graph \(g\) represents its derivative. Thus, \(g = f'\).

- 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 blue graph \(k_1\) has a local maximum at \(x = 1\) and a local minimum at \(x = 3\). The red graph \(k_2\) 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) \(f(2) = 2\) \(f'(2) = -1.5\) \(f(3) = 1\) \(f'(1) = 0\)

- 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 \(f''(x)<0\).

- How is \(f\) related to the slope of \(F\)? - Use zeros and sign changes of \(f\) to identify extrema of \(F\). - Use the sign of \(f\) to determine where \(F\) increases or decreases. - What feature of \(f\) corresponds to an inflection point of \(F\)?

1. Since \(F'=f\), zeros and signs of \(f\) determine the critical points and monotonicity of \(F\). Also, extrema of \(f\) correspond to inflection points 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 \(F\) has an inflection point there. Statement c is true.

a) True b) False c) True

- How is the derivative \(f\) related to the monotonicity of \(F\)? - Look for zeros of \(f\) with sign changes. - Use the direction of each sign change to classify the extremum. - What feature of \(f\) corresponds to an inflection point of \(F\)?

1. Since \(F'=f\), 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. Inflection points of \(F\) occur at extrema of \(f\). The graph of \(f\) has a local minimum at \(x=1\), so \(F\) has an inflection point there.

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 red graph \(g_1\) has horizontal tangents at \(x = -2\) and \(x = 2\), while the blue graph \(g_2\) 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 \((2, 5)\)

- 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) \(-0.75\)

- 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)

- 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

- Identify where the graph of \(f'\) is increasing and decreasing. - How does an increasing or decreasing first derivative determine concavity? - Inflection values 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 inflection values 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>

- 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 values \(x=-1\) 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 \(x_5\)

- How does the sign of the second derivative determine concavity? - An inflection value 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 inflection values 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 \(f''(0)=-4\ne0\).

- 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-point 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 points and compare their locations and classifications.

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 points 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 values 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 \([-5.2, -3)\) and \((-1, 0.6]\), 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 inflection values 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 \([-5.2, -3)\) and \((-1, 0.6]\); 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 \(1\)

- A cubic has four coefficients, and the two extrema provide four equations. - At each extremum, use both the point and zero-slope conditions. - Substitute \(x=2\) to test the student's point. - Find the midpoint of the two extrema.

1. 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.

- How is the slope of an antiderivative related to the original function? - Compare extrema of one graph with zeros of the other. - Compare increasing and decreasing intervals with the sign of the other graph. - Check whether all these features agree.

1. The slope of \(F\) at each x-value equals \(f(x)\). 2. Graph a has a local maximum at \(x=-2\) and a local minimum at \(x=2\). Therefore, its derivative must be zero at those x-values. Graph b has zeros at \(x=-2\) 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 the antiderivative \(F\), and graph b represents \(f\).

Graph a represents \(F\), and graph b represents \(f\).

- The sign of a function determines whether its antiderivative increases or decreases. - A sign-changing zero corresponds to a local extremum of an antiderivative. - Identify the zeros of \(f\) and the required extrema of its antiderivative. - Repeat the same analysis for \(g\).

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, an antiderivative of \(f\) 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, an antiderivative of \(g\) must increase before \(x=1\), decrease afterward, and have a local maximum at \(x=1\). Candidate b has this behavior.

\(f\) matches candidate a. \(g\) matches candidate b.

- Use the sign of \(f\) to determine where \(F\) increases or decreases. - A sign change of \(f\) identifies and classifies a local extremum of \(F\). - Extrema of \(f\) correspond to inflection points of \(F\). - Interpret \(F(3)-F(1)\) as a definite integral.

1. Statement (1) is true. The graph of \(f\) is positive on \((-2,1)\), so \(F\) is strictly increasing on \([-2,1]\). 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 \(x=2.1\), so \(F\) has an inflection point there. 4. Statement (4) is false. Since \(f(x)<0\) on \((1,3)\), \(F(3)-F(1)=\int_1^3 f(x)\,dx<0\). Therefore, \(F(3)<F(1)\).

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

- Zeros of \(f\) correspond to critical points of an antiderivative. - The sign of \(f\) determines whether the antiderivative increases or decreases. - Compare the extrema of each candidate with the zeros of \(f\). - Check the monotonicity before the first zero.

1. The graph of \(f\) has zeros at \(x=1\) and \(x=3\). Therefore, an antiderivative \(F\) must have local extrema at those x-values. 2. Candidate 2 has extrema at \(x=0\) and \(x=2\), so it cannot be an antiderivative of \(f\). 3. For \(x<1\), \(f(x)>0\), so \(F\) must be increasing. Candidate 1 increases before \(x=1\) and has a local maximum there. 4. Candidate 3 decreases before \(x=1\) and has a local minimum there, which contradicts \(f(x)>0\). 5. Therefore, candidate 1 is an antiderivative of \(f\).

a) Candidate 1 b) Candidate 2 has extrema at the wrong x-values, and candidate 3 has the opposite monotonicity and extremum types.

- Use zeros and sign changes of \(f\) to identify extrema of \(F\). - Integrate the quadratic function term by term. - Use the point \((0,1)\) to determine \(C\). - Substitute the critical x-values into \(F\).

1. The zeros of \(f\) are \(x=-3\) and \(x=3\). At \(x=-3\), \(f\) 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. 2. Integrate: \(F(x)=-\frac{1}{18}x^3+1.5x+C\). 3. Since \(F(0)=1\), \(C=1\). Thus, \(F(x)=-\frac{1}{18}x^3+1.5x+1\). 4. Evaluate the critical points: \(F(-3)=-2\) and \(F(3)=4\). 5. Therefore, the local minimum is \((-3,-2)\), and the local maximum is \((3,4)\), as predicted.

a) Local minimum at \(x=-3\); local maximum at \(x=3\) b) \(F(x)=-\frac{1}{18}x^3+1.5x+1\) c) Local minimum \((-3,-2)\); local maximum \((3,4)\)

- The sign of \(f\) determines the monotonicity of \(F\). - Integrate the polynomial and use the given point to find \(C\). - What feature of \(f\) corresponds to an inflection point of \(F\)? - Check whether the slope of \(f\) changes sign at \(x=0\).

1. Since \(F'=f\), \(F\) is strictly increasing on \((-2, 2)\), where \(f>0\), and strictly decreasing on \((-\infty, -2)\) and \((2, \infty)\), where \(f<0\). 2. The general antiderivative is \(F(x)=-\frac{1}{3}x^3+4x+C\). 3. Use \(F_1(3)=2\): \(-9+12+C=2\), so \(C=-1\). Therefore, \(F_1(x)=-\frac{1}{3}x^3+4x-1\). 4. Yes. The graph of \(f\) has a local maximum at \(x=0\), so the slope of \(f\) changes from positive to negative. Since \(F''=f'\), the concavity of \(F\) changes at \(x=0\).

a) Increasing on \((-2, 2)\); decreasing on \((-\infty, -2)\) and \((2, \infty)\) b) \(F_1(x)=-\frac{1}{3}x^3+4x-1\) c) Yes, every antiderivative has an inflection point at \(x=0\).

- 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, extrema, and monotonicity among the three graphs. - The slope of \(F\) equals the value of \(f\). - Extrema of \(f\) correspond to zeros of \(f'\). - Compare polynomial degrees to find the end behavior of the rational function.

1. Graph b represents \(f\), because it passes through the origin and has values \(f(1)=\frac{3}{2}\) and \(f(-1)=-\frac{3}{2}\). 2. Graph a represents \(F\). Since \(f<0\) for \(x<0\) and \(f>0\) for \(x>0\), an antiderivative decreases to a local minimum at \(x=0\) and then increases. 3. Graph c represents \(f'\). The function \(f\) has extrema at \(x=\pm1\), so \(f'\) has zeros there. 4. The slope of \(F\) at \(x=2\) is \(F'(2)=f(2)=\frac{6}{5}\). 5. Differentiating gives \(f'(x)=\frac{3-3x^2}{(x^2+1)^2}\). The denominator has higher degree than the numerator, so \(\lim_{x\to\pm\infty}f'(x)=0\). The horizontal asymptote is \(y=0\).

a) a: \(F\); b: \(f\); c: \(f'\) b) \(\frac{6}{5}\) c) \(\lim_{x\to\pm\infty}f'(x)=0\); horizontal asymptote \(y=0\)

- Use the sign of \(f\) to determine the monotonicity of \(F\). - Classify a critical point of \(F\) from the sign change of \(f\). - The increase or decrease of \(f\) determines the concavity of \(F\). - Extrema of \(f\) correspond to inflection points of \(F\).

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. Therefore, \(f'(x)>0\), so \(F''(x)>0\), and \(F\) is concave up. Statement (3) is true. 4. Inflection points of \(F\) correspond to extrema of \(f\). 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 zeros and sign changes of \(f\) to classify extrema of \(F\). - Write the quadratic in vertex form. - Use a zero to determine the leading coefficient. - Integrate and apply \(F(0)=2\).

1. Since \(F'=f\), the zeros of \(f\) are critical points of \(F\). At \(x=0\), \(f\) 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. 2. Use vertex form: \(f(x)=a(x-1)^2-1.5\). Since \(f(0)=0\), \(a-1.5=0\), so \(a=1.5\). 3. Therefore, \(f(x)=1.5(x-1)^2-1.5=1.5x^2-3x\). 4. An antiderivative is \(F(x)=0.5x^3-1.5x^2+C\). Since \(F(0)=2\), \(C=2\). 5. Thus, \(F(3)=0.5(27)-1.5(9)+2=2\).

a) Local maximum at \(x=0\); local minimum at \(x=2\) b) \(f(x)=1.5x^2-3x\), and \(F(3)=2\)

- 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

- At a local maximum or minimum, the derivative is \(0\). - Compare where one graph increases or decreases with where another graph is positive or negative. - Build a derivative chain among the three graphs.

1. Graph b has a local maximum at \(x=0\). Its derivative must be \(0\) 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, is positive where graph c increases, and is negative where graph c decreases. 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 \(f^{\prime}\).

Graph a: \(f^{\prime}\) Graph b: \(F\) Graph c: \(f\)

- 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 \(a=\frac{1}{4}\) and \(c=-3\). Therefore, \(f(x)=\frac{1}{4}x^3-3x\).

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

- Zeros of \(f\) correspond to horizontal tangents of an antiderivative. - Where \(f>0\), an antiderivative increases; where \(f<0\), it decreases.

1. The function \(f\) has zeros at \(x=-2\) and \(x=2\), so an antiderivative 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 an antiderivative must be decreasing there. Graph \(g\) decreases on that interval, while graph \(h\) increases. 3. Therefore, graph \(g\) is an antiderivative of \(f\).

Graph \(g\) is an antiderivative of \(f\).

- 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 \(p'(2) \approx -1.8\)

- 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 point 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

- 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

- 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: \(f'(x)\) 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

- 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

- 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

- Observe how the negative tangent slopes change from left to right. - Do the slopes become more negative or approach zero? - How does concavity relate to whether the derivative increases or decreases?

1. The first derivative is \(f'(x)=4\ln(0.75)(0.75)^x\). Since \(\ln(0.75)<0\), the function \(f\) is strictly decreasing and \(f'(x)<0\) for every \(x\). 2. The second derivative is \(f''(x)=4[\ln(0.75)]^2(0.75)^x>0\). Therefore, the graph of \(f\) is concave up. 3. Because \(f''(x)>0\), the derivative \(f'\) is strictly increasing. Its graph remains below the x-axis and approaches the x-axis from below as \(x\to\infty\).

The graph of \(f\) is concave up. The derivative \(f'\) is strictly increasing, remains negative, and approaches \(0\) from below as \(x\to\infty\).

- 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 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 \(x = \pm 1\). Thus, C matches graph 3.

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

- 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)

- 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 \((0.5, 0.4)\)

- 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 \(x=2\).

- 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.

- Use the sign of \(f^{\prime}\) to determine where \(f\) increases and decreases. - A derivative that is \(0\) without changing sign does not create an extremum. - Turning points of \(f^{\prime}\) indicate changes in the concavity of \(f\). - Use the given point to locate the graph vertically.

1. The derivative is positive on \((-3,0)\) except at \(x=-2\), where it is \(0\) without changing sign. Thus, \(f\) increases through \(x=-2\). The graph of \(f^{\prime}\) has a local minimum at \(x=-2\), so \(f\) has a stationary inflection point there. The graph of \(f^{\prime}\) then has a local maximum at approximately \(x=-0.7\), so \(f\) has another inflection point there. At \(x=0\), \(f^{\prime}\) changes from positive to negative, so \(f\) has a local maximum. Because \(f\) passes through \((0,0)\), that local maximum is \((0,0)\). 2. On \([-3,0]\), \(f\) is increasing, and on \((0,1]\), it is decreasing. Therefore, the greatest value on \([-3,1]\) occurs at \(x=0\).

1. The graph passes through \((0,0)\), has a stationary inflection point at \(x=-2\), another inflection point at approximately \(x=-0.7\), and a local maximum at \((0,0)\). 2. \(x=0\)

- 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 inflection values 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 \((0, 2)\)

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

- 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.

- 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.

- 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 \(g(0) \approx -1.2\).

- 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 \(f''\). - Inflection values of \(f\) occur at local extrema of \(f'\). - 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 \((2, 4.5]\).

- 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 inflection values. 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 values: \(x=-4\), \(x=0\), and \(x=4\). Concave down on \([-6, -4)\) and \((0, 4)\).

- An inflection point of \(f\) occurs where \(f^{\prime}\) changes from increasing to decreasing or vice versa. - The signed area under \(f^{\prime}\) gives the change in \(f\). - Use the sign of \(f^{\prime}\) for monotonicity and its increasing or decreasing behavior for concavity.

1. The derivative \(f^{\prime}\) changes from increasing to decreasing at \(x=2\). Therefore, the concavity of \(f\) changes there, so \(f\) has an inflection point at \(x=2\). 2. By accumulation of change, \(f(4)-f(0)=\int_0^4 f^{\prime}(x)\,dx\). The region is a triangle with base \(4\) and height \(2\), so its area is \(\frac12\cdot4\cdot2=4\). Thus, \(f(4)=4\). 3. From \(x=0\) to \(x=4\), \(f^{\prime}>0\), so \(f\) increases from \((0,0)\) to a local maximum at \((4,4)\). The area from \(0\) to \(2\) is \(2\), so the inflection point is \((2,2)\). The graph is concave up on \((0,2)\), concave down on \((2,6)\), and decreases after \(x=4\).

a) \(x=2\) b) \(f(4)=4\) c) The graph passes through \((0,0)\), has an inflection point at \((2,2)\), reaches a local maximum at \((4,4)\), and then decreases. It is concave up on \((0,2)\) and concave down on \((2,6)\).

- 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 \(C\): \(f(x)>0\), \(f'(x)<0\), \(f''(x)=0\)

- 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 value 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 an inflection value. 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

- Use the sign of \(f''\) to determine concavity. - Use \(f''\) as the derivative of \(f'\) to determine whether \(f'\) increases or decreases. - What must happen to \(f''\) at an inflection value? - 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

- Which derivative determines the concavity of a function? - 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. 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 a sufficient condition for an inflection point. Therefore, all three values are inflection values. 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 \(x=3.5\). The equation \(f''(x)=0\) is necessary here, 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 values 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 value. 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: 1. \(f'\) must be cubic, but \(g\) is quadratic. 2. \(f'\) must be odd, but \(g\) is even. 3. \(f'(0)=0\), 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: 1. \(f'\) must be cubic, but \(g\) is quadratic. 2. \(f'\) has a double zero and no sign change at \(x=0\), but \(g\) changes sign there. 3. \(f'(x)<0\) 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. - Inflection values of \(f\) occur at local extrema of \(f'\). - 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 \(3<x\le5.5\).

- What must the second derivative do at an inflection value? - 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.

- 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 \(h(x)=f(x-2)+1=\frac{1}{4}((x-2)^3-9(x-2))+1\).

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 inflection values 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. Inflection values 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 \(f(-1)>f(-2)\).

- Use \(F^{\prime}=f\) to connect the sign of \(f\) to the monotonicity of \(F\). - Local extrema of \(F\) require a sign change of \(f\) through \(0\). - Compare two values of \(F\) using a definite integral of \(f\).

1. True. On \([-3,-1]\), \(f\ge0\), and \(f>0\) in the interior. Since \(F^{\prime}=f\), \(F\) is strictly increasing there. 2. False. Near \(x\approx0.8\), the graph of \(f\) has a local minimum but is not \(0\). Therefore, \(F^{\prime}\ne0\), so \(F\) does not have a local extremum there. Instead, \(F\) has an inflection point because \(f\) changes from decreasing to increasing. 3. True. At \(x=-3\), \(f\) changes from negative to positive, so \(F^{\prime}\) changes from negative to positive. Thus, \(F\) has a local minimum. 4. False. By the Fundamental Theorem of Calculus, \(F(2)-F(-1)=\int_{-1}^{2}f(x)\,dx\). The graph lies below the x-axis on \((-1,2)\), so this integral is negative. Therefore, \(F(2)<F(-1)\).

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

- The sign of \(f\) determines whether \(F\) increases or decreases. - A sign change of \(f\) determines the type of local extremum of \(F\). - The increasing or decreasing behavior of \(f\) determines the concavity of \(F\). - The value \(f(x)\) is the slope of \(F\) at that x-value.

1. True. The graph shows \(f>0\) on \([1,4]\). Since \(F^{\prime}=f\), \(F\) is strictly increasing there. 2. True. At \(x=5\), \(f\) changes from positive to negative, so \(F\) has a local maximum. 3. False. The function \(f\) is increasing on \((0,2)\), so \(F^{\prime}=f\) is increasing and \(F\) is concave up there. 4. True. The slope is \(F^{\prime}(3)=f(3)=2\).

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

- Compare the value of one graph at \(x=0\) with the slope of the other graph there. - Check whether the sign and size of one graph agree with the direction and steepness of the other. - Use \(F^{\prime}=f\).

1. Since \(F^{\prime}=f\), the value of \(f\) at any x-value must equal the slope of \(F\) there. 2. At \(x=0\), graph (2) has value \(0.5\). Graph (1) has value \(1\) and a positive slope of about \(0.5\), so graph (2) matches the slope of graph (1) 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 \(F\), and graph (2) is \(f\).

- If \(f\) is positive, what must an antiderivative do? - Compare the size of one graph with the steepness of the other. - Check the relationship near \(x=0\).

1. Graph (1) is positive throughout the displayed interval. If graph (1) is \(f\), then an antiderivative must be strictly increasing. Graph (2) has that behavior. 2. Graph (1) decreases toward \(0\) as \(x\) increases across the display. Accordingly, the positive slope of graph (2) should get smaller, so graph (2) should become flatter. The graph shows this behavior. 3. Near \(x=0\), graph (1) has value about \(1\), and graph (2) has slope about \(1\), which is consistent with \(F^{\prime}=f\). 4. Therefore, graph (1) is \(f\), and graph (2) is \(F\).

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

- 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 \(g'\).

- An antiderivative of a quadratic function is cubic. - Zeros and sign changes of \(f\) determine local extrema of \(F\). - Use \(F(0)=2\) to position the graph vertically.

1. The graph represents \(f(x)=x^2-4\), so an antiderivative is \(F(x)=\frac13x^3-4x+C\). 2. Since \(F(0)=2\), \(C=2\). Thus, one required antiderivative is \(F(x)=\frac13x^3-4x+2\). 3. The function \(f\) changes from positive to negative at \(x=-2\), so \(F\) has a local maximum there. It changes from negative to positive at \(x=2\), so \(F\) has a local minimum there. 4. The function \(f\) has a local minimum at \(x=0\), so \(F\) changes concavity there. Therefore, \((0,2)\) is an inflection point of \(F\).

\(F(x)=\frac13x^3-4x+2\); the graph has a local maximum at \(x=-2\), a local minimum at \(x=2\), and an inflection point at \((0,2)\).

- Compare the y-intercept and end behavior of \(G\) with the four graphs. - Zeros of \(g=G^{\prime}\) occur at horizontal tangents of \(G\). - The derivative of an even function is odd. - Use the monotonicity of \(G\) to determine the sign of \(g\).

1. Since \(G(0)=2\) and \(G(x)\to0\) as \(x\to\pm\infty\), graph (1) represents \(G\). Possible equations for the others are graph (2): \(y=\frac{4}{x^2+2}-3\), graph (3): \(y=-\frac{4}{x^2+2}\), and graph (4): \(y=\frac{3x}{x^2+1}\). 2. Since \(g=G^{\prime}\), zeros of \(g\) occur where \(G\) has horizontal tangents. Graph (1) has exactly one such point, its maximum at \(x=0\). Therefore, \(g\) has exactly one zero at \(x=0\). 3. The function \(G\) is even, so its derivative \(g\) is odd. Thus, the graph of \(g\) is symmetric about the origin. 4. Differentiating to confirm the qualitative description gives \(g(x)=-\frac{8x}{(x^2+2)^2}\). Hence \(g(x)>0\) for \(x<0\), \(g(0)=0\), and \(g(x)<0\) for \(x>0\). Also, \(g(x)\to0\) as \(x\to\pm\infty\), so the horizontal asymptote is \(y=0\).

a) Graph (1). Possible equations: (2) \(y=\frac{4}{x^2+2}-3\), (3) \(y=-\frac{4}{x^2+2}\), (4) \(y=\frac{3x}{x^2+1}\) b) One zero at \(x=0\) c) \(G\) is even; \(g\) is odd. d) Horizontal asymptote \(y=0\); \(g>0\) for \(x<0\), \(g(0)=0\), and \(g<0\) for \(x>0\)

- Use \(F^{\prime}=f\). - A sign change of \(f\) determines a local extremum of \(F\). - The graph of \(F\) is concave down where \(f\) is decreasing. - Consider both the limiting slope and the remaining accumulated area.

1. At \(x=1\), \(f=F^{\prime}\) changes from negative to positive, so \(F\) has a local minimum there. 2. The graph of \(F\) is concave down where \(F^{\prime}=f\) is decreasing. The graph of \(f\) decreases for \(x>3\), so \(F\) is concave down there. 3. As \(x\to\infty\), \(f(x)\to0\), so the slope \(F^{\prime}(x)\to0\). The formula for \(f\) has exponential decay, so the remaining accumulated area is finite; this indicates that \(F\) approaches a horizontal asymptote.

a) Local minimum at \(x=1\) b) Concave down for \(x>3\) c) The slope approaches \(0\), and \(F\) approaches a horizontal asymptote.

- 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. - Signed area under \(f^{\prime}\) gives the change in \(f\).

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. The change is \(f(5)-f(0)=\int_0^5 f^{\prime}(x)\,dx\). The negative signed area from \(0\) to \(4\) has greater magnitude than the positive area from \(4\) to \(5\), so the net change is negative. Therefore, \(f(5)<4\).

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) \(f(5)<4\)

- Use zeros and sign changes of \(f'\) to locate and classify local extrema of \(f\). - Use the sign of \(f'\) to determine where \(f\) increases or decreases. - Include the given point \((0, 1)\) in your description. - An inflection point of \(f\) corresponds to a local extremum of \(f'\).

1. The zeros of \(f'\) are \(x=-1\) and \(x=3\). At \(x=-1\), \(f'\) 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. 2. The graph of \(f\) increases for \(x<-1\), decreases for \(-1<x<3\), and increases for \(x>3\). It must pass through \((0, 1)\). 3. The graph of \(f'\) has a local minimum at \(x=1\), where it changes from decreasing to increasing. Therefore, the concavity of \(f\) changes at \(x=1\), so \(f\) has an inflection point there.

1. Local maximum at \(x=-1\); local minimum at \(x=3\). 2. The graph increases for \(x<-1\), decreases for \(-1<x<3\), increases for \(x>3\), and passes through \((0, 1)\). 3. The inflection point occurs at \(x=1\).

- 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) \(f_4\)

- The value of \(f\) is the slope of \(F\). - Horizontal tangents of \(F\) give zeros of \(f\). - Concavity of \(F\) determines the sign of \(f^{\prime}\). - Use \(\int_a^bf(x)\,dx=F(b)-F(a)\).

1. Since \(f=F^{\prime}\), zeros of \(f\) occur where \(F\) has horizontal tangents. The graph of \(F\) has local extrema at \(x=-2\) and \(x=2\), so these are the zeros of \(f\). 2. The function \(f\) is negative where \(F\) is decreasing, which is \((-2,2)\). 3. Since \(f^{\prime}=F^{\prime\prime}\), \(f^{\prime}>0\) where \(F\) is concave up. The graph changes concavity at \(x=0\) and is concave up for \(x>0\), so the graph of \(f^{\prime}\) is above the x-axis on \((0,\infty)\). 4. By the Fundamental Theorem of Calculus, \(\int_{-2}^{2}f(x)\,dx=F(2)-F(-2)=-4-4=-8\). 5. At \(x=-3\), \(F\) is increasing, so \(f(-3)>0\). At \(x=0\), \(F\) is decreasing, so \(f(0)<0\). Therefore, \(f(-3)>f(0)\).

a) \(x=-2\) and \(x=2\) b) \((-2,2)\) c) \((0,\infty)\) d) \(-8\) e) True

- The value of \(f\) is the slope of \(F\). - Use where \(F\) increases or decreases to determine the sign of \(f\). - Apply the Fundamental Theorem of Calculus. - The minimum of a quadratic lies midway between its zeros.

1. Zeros of \(f=F^{\prime}\) occur at local extrema of \(F\). The graph has a local maximum at \(x=1\) and a local minimum at \(x=3\), so the zeros are \(1\) and \(3\). 2. The function \(f\) is nonnegative where \(F\) is increasing: \((-\infty, 1]\cup[3, \infty)\). 3. By the Fundamental Theorem of Calculus, \(\int_1^4f(x)\,dx=F(4)-F(1)=4-4=0\). 4. The quadratic \(f\) has zeros at \(1\) and \(3\) and is positive outside those zeros, so it opens upward. Its minimum occurs midway between the zeros at \(x=2\). This is also the inflection-point x-value of \(F\), where the slope of \(F\) is smallest. 5. Since \(F\) is increasing at \(x=0\), \(f(0)>0\). Since \(F\) is decreasing at \(x=2\), \(f(2)<0\).

a) \(x=1\) and \(x=3\) b) \((-\infty, 1]\cup[3, \infty)\) c) \(0\) d) \(x=2\) e) \(f(0)>0\) and \(f(2)<0\)

- A negative-to-positive sign change of \(f\) gives a local minimum of \(F\). - Local extrema of \(f\) correspond to inflection points of \(F\). - A vertical shift does not affect where a function increases or decreases.

1. A local minimum of \(F\) occurs where \(f=F^{\prime}\) changes from negative to positive. The graph shows this at \(x=-4\) and \(x=3\). 2. Inflection points of \(F\) occur at local extrema of \(f\). 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 \(F^{\prime}=f>0\). The value \(F(0)\) only changes the vertical position of \(F\), not its monotonicity.

a) \(x=-4\) and \(x=3\) b) Approximately \(x=-2.7\) and \(x=1.4\) c) True

- Zeros of a derivative correspond to horizontal tangents of the original function. - Compare signs with increasing and decreasing intervals. - Build a derivative chain among the three graphs.

1. Graph 1 has local extrema at \(x=-1\) and \(x=1\). Graph 2 has zeros at those x-values and signs that match the increasing and decreasing behavior of graph 1. Therefore, graph 2 is the derivative of graph 1. 2. Graph 1 has a zero at \(x=0\), changing from negative to positive. Graph 3 has a local minimum at \(x=0\). Therefore, graph 1 is the derivative of graph 3. 3. Hence graph 1 is \(f\), graph 2 is \(f^{\prime}\), and graph 3 is \(F\).

Graph 1: \(f\) Graph 2: \(f^{\prime}\) Graph 3: \(F\)

- Count the graph's turning points and relate that number to polynomial degree. - Decide whether the graph has y-axis symmetry or \(180^\circ\) rotational symmetry about the origin. - For origin symmetry, keep only odd powers in the cubic form. - Use both the coordinates and the horizontal tangent at the marked local maximum.

1. The graph has two local extrema. A polynomial of degree \(n\) has at most \(n-1\) turning points, so its degree is at least \(3\). A cubic is therefore the least possible degree. 2. The graph has \(180^\circ\) rotational symmetry about the origin, so \(f\) is odd. For a cubic, \(f(x)=ax^3+cx\). 3. The marked local maximum is \((-1, 2)\). Therefore, \(f(-1)=2\) and \(f'(-1)=0\). These conditions give \(-a-c=2\) and \(3a+c=0\). Solving the system gives \(a=1\) and \(c=-3\). Thus, \(f(x)=x^3-3x\).

1. Least possible degree: \(3\) 2. \(f(x)=ax^3+cx\) 3. \(f(x)=x^3-3x\)

- Count the graph's turning points and relate that number to polynomial degree. - Compare the left and right halves of the graph to identify its symmetry. - Determine which powers can appear in a polynomial that is symmetric about the y-axis. - Use both the coordinates and the horizontal tangent at one marked extremum.

1. The graph has three local extrema. A polynomial of degree \(n\) has at most \(n-1\) turning points, so the degree must be at least \(4\). 2. The graph is symmetric about the y-axis, so \(f\) is even. For a fourth-degree polynomial, write \(f(x)=ax^4+bx^2+c\). 3. The marked local maximum \((0, 0)\) gives \(c=0\), so \(f(x)=ax^4+bx^2\). The local minimum \((2, -4)\) gives \(16a+4b=-4\). Because the tangent there is horizontal, \(f'(2)=0\). Since \(f'(x)=4ax^3+2bx\), this gives \(32a+4b=0\). Solving the system yields \(a=\frac{1}{4}\) and \(b=-2\). Therefore, \(f(x)=\frac{1}{4}x^4-2x^2\).

1. The degree is at least \(4\) because the graph has three turning points. 2. The graph is symmetric about the y-axis, so \(f(x)=ax^4+bx^2+c\). 3. \(f(x)=\frac{1}{4}x^4-2x^2\)

- Extrema of a function occur at sign-changing zeros of its derivative. - The sign of a derivative determines increasing and decreasing intervals. - Test pairs of graphs systematically. - Use zeros, extrema, and monotonicity together.

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 has signs consistent with the increasing and decreasing intervals of graph b. Thus, graph a is the derivative of graph b. 2. Graph b has zeros at \(x=-\sqrt{3}\), \(x=0\), and \(x=\sqrt{3}\). Graph c has local extrema at those x-values, with monotonicity matching 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 monotonicity opposite to what graph a would require as its derivative, so it cannot complete the derivative-antiderivative chain.

Graph c represents \(F\). Graph b represents \(f\). Graph a represents \(f'\). Graph d does not belong to the group.

- Compare the domains and look for vertical asymptotes. - Identify zeros and y-intercepts. - Differentiate each candidate to locate critical points 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 points 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: \(f_2(x)=x^2e^{-x}\)

- Use the zeros to write the quadratic in factored form. - How are zeros of \(f\) related to critical points of an antiderivative \(F\)? - Integrate the polynomial term by term. - Use the point \(A\) to determine the constant.

1. The graph shows zeros at \(x=-1\) and \(x=3\), and a vertex at \((1,4)\). Write \(f(x)=a(x+1)(x-3)\). 2. Use the vertex: \(4=a(2)(-2)=-4a\), so \(a=-1\). Thus, \(f(x)=-x^2+2x+3\). 3. The general antiderivative is \(F(x)=-\frac{1}{3}x^3+x^2+3x+C\). 4. Use \(F(3)=5\): \(-9+9+9+C=5\), so \(C=-4\). Therefore, \(F(x)=-\frac{1}{3}x^3+x^2+3x-4\). 5. Since \(F'=f\), the zeros of \(f\) are critical points of \(F\). At \(x=-1\), \(f\) changes from negative to positive, so \(F\) has a local minimum with value \(F(-1)=-\frac{17}{3}\). At \(x=3\), \(f\) changes from positive to negative, so \(F\) has a local maximum with value \(F(3)=5\). 6. Since \(F''=f'\), the vertex of \(f\) at \(x=1\) corresponds to an inflection point of \(F\), with \(F(1)=-\frac{1}{3}\). Also, \(F(4)=\frac{8}{3}\). These features describe the graph on \([-1,4]\).

a) \(f(x)=-x^2+2x+3\) b) \(F(x)=-\frac{1}{3}x^3+x^2+3x-4\) c) On \([-1,4]\), \(F\) has a local minimum at \(\left(-1,-\frac{17}{3}\right)\), an inflection point at \(\left(1,-\frac{1}{3}\right)\), a local maximum at \((3,5)\), and endpoint values \(F(-1)=-\frac{17}{3}\) and \(F(4)=\frac{8}{3}\).

- Use \(F^{\prime}=f\) to determine where \(F\) increases or decreases. - Check sign changes of \(f\) at its zeros. - Local extrema of \(f\) correspond to inflection points of \(F\). - Use factored form from the zeros and one additional point to determine the leading coefficient.

1. Since \(F^{\prime}=f\), \(F\) is strictly decreasing where \(f<0\). The cubic is negative on \((-\infty, -4)\) and \((1, 5)\). 2. At \(x=-4\), \(f\) 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. At \(x=5\), \(f\) changes from negative to positive, so \(F\) has a local minimum. 3. To obtain the exact inflection-point x-values in part c, first use the cubic form requested in part d. 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 \(a=0.1\). Thus, \(f(x)=0.1(x+4)(x-1)(x-5)\). 4. Inflection points of \(F\) occur at local extrema of \(f\). Differentiating the equation above gives \(f^{\prime}(x)=0.1(3x^2-4x-19)\). Thus, the x-values are \(x=\frac{2-\sqrt{61}}{3}\approx-1.94\) and \(x=\frac{2+\sqrt{61}}{3}\approx3.27\).

a) \((-\infty, -4)\) and \((1, 5)\) b) Local minima at \(x=-4\) and \(x=5\); local maximum at \(x=1\) c) \(x=\frac{2-\sqrt{61}}{3}\approx-1.94\) and \(x=\frac{2+\sqrt{61}}{3}\approx3.27\) d) \(f(x)=0.1(x+4)(x-1)(x-5)\)

- Local extrema of \(f\) correspond to inflection points of an antiderivative. - Zeros and signs of \(f\) determine the local extrema and monotonicity of \(F_1\). - Use the zeros and y-intercept to determine the quadratic. - Geometric area is positive even when the definite integral is negative.

1. Since \(F^{\prime}=f\), inflection points of \(F\) occur at local extrema of \(f\). The graph of \(f\) has exactly one local extremum on \([-2,2]\), a minimum at \(x=0\). Therefore, every antiderivative has exactly one inflection point there. 2. To determine \(F_1\) and solve part c, first determine \(f\). The zeros are \(x=\pm2\), so \(f(x)=a(x-2)(x+2)\). Since the graph shows \(f(0)=-2\), \(-4a=-2\), giving \(a=\frac12\). Thus, \(f(x)=\frac12x^2-2\). 3. Integrating gives \(F_1(x)=\frac16x^3-2x+C\). The condition \(F_1(2)=0\) gives \(C=\frac83\). Thus, \(F_1(x)=\frac16x^3-2x+\frac83\). Its graph has a local maximum at \((-2,\frac{16}{3})\), an inflection point at \((0,\frac83)\), and a local minimum at \((2,0)\). 4. The graph of \(f\) is below the x-axis on \([-2,2]\), so the area is \(-\int_{-2}^{2}(\frac12x^2-2)\,dx=\frac{16}{3}\).

a) The only inflection-point x-value is \(x=0\). b) \(F_1(x)=\frac16x^3-2x+\frac83\); local maximum at \((-2,\frac{16}{3})\), inflection point at \((0,\frac83)\), and local minimum at \((2,0)\) c) \(f(x)=\frac12x^2-2\); area \(=\frac{16}{3}\) square units

- 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) \(f^{\prime}(x)=2\cos(\frac{\pi}{3}x)\)

- Use \(F^{\prime}=f\). - Zeros and sign changes of \(f\) determine local extrema of \(F\). - Local extrema of \(f\) determine inflection points of \(F\). - Interpret changes in \(F\) as accumulated signed area under \(f\).

1. True. At \(x=0\), \(f=F^{\prime}\) changes from negative to positive, so \(F\) has a local minimum. 2. False. On \((0,3)\), \(f>0\), so \(F\) is strictly increasing. 3. False. Inflection points of \(F\) occur at local extrema of \(f\). At \(x=0\), \(f\) is increasing through a zero rather than having a local extremum. 4. True. For \(x>3\), \(f<0\) and decreases without bound. The accumulated signed area from any fixed point becomes arbitrarily negative, so \(F(x)\to-\infty\).

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

- 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 \(x=\pm1\).

- Zeros of a derivative correspond to horizontal tangents of the original function. - Compare the signs of a possible derivative with where the original graph increases or decreases. - A sign change in a derivative can identify a local extremum of the original function. - Apply the derivative relationship repeatedly to connect all four graphs.

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. Therefore, Graph 4 represents \(F\), and Graph 3 represents \(f\). 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. Therefore, Graph 1 represents \(f^{\prime}\). 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. Therefore, Graph 2 represents \(f^{\prime\prime}\).

Graph 1: \(f^{\prime}\) Graph 2: \(f^{\prime\prime}\) Graph 3: \(f\) Graph 4: \(F\)

- 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: \(g_3(x)=2-e^{0.5x}\)