Free Math Worksheets

- How is concavity related to the sign of the second derivative? - Why can the expression for the second derivative never be zero or negative? - What property of squares is useful here?

1. Differentiate twice: \(g'(x)=x^3+6x\) and \(g''(x)=3x^2+6\). 2. For every real number \(x\), \(x^2 \ge 0\). Therefore, \(3x^2+6 \ge 6>0\). 3. Since \(g''(x)>0\) for all \(x \in \mathbb{R}\), the graph of \(g\) is concave up on \((-\infty, \infty)\).

Because \(g''(x)=3x^2+6 \ge 6>0\) for all real \(x\), the graph is concave up on \((-\infty, \infty)\).

- What does the sign of the second derivative tell you about the graph? - Which conditions make the second derivative test conclusive? - What derivative condition guarantees a change in concavity? - Decide whether each condition concerns slope, concavity, or change in concavity.

1. Condition 1 is the second derivative test for a local minimum, so 1 matches C. 2. Condition 2 states that the second derivative is negative throughout \(I\), so the graph is concave down there. Thus, 2 matches A. 3. Condition 3 guarantees that the second derivative changes sign at \(x_0\), so the graph has an inflection point. Thus, 3 matches B.

1. C 2. A 3. B

- Which derivative describes concavity? - What necessary equation must hold at an inflection point? - How can the third derivative confirm a change in concavity? - Remember that a quadratic equation can have two real solutions.

1. Differentiate: \(f'(x)=\frac{1}{3}x^3-2x\), \(f''(x)=x^2-2\), and \(f'''(x)=2x\). 2. Solve \(f''(x)=0\): \(x^2-2=0\), so \(x=-\sqrt{2}\) and \(x=\sqrt{2}\). 3. Since \(f'''(-\sqrt{2})=-2\sqrt{2}\ne0\) and \(f'''(\sqrt{2})=2\sqrt{2}\ne0\), the second derivative changes sign at both values.

\(x=-\sqrt{2}\) and \(x=\sqrt{2}\)

- Which derivative determines concavity? - What signs of that derivative correspond to concave up and concave down? - What does the second derivative tell you about the first derivative? - How does a function behave when its derivative is positive? Apply that idea to \(f'\).

1. Differentiate: \(f'(x)=3x^2-12x+10\) and \(f''(x)=6x-12\). 2. Solve \(f''(x)=0\): \(6x-12=0\), so \(x=2\). 3. For \(x<2\), \(f''(x)<0\), so the graph is concave down. For \(x>2\), \(f''(x)>0\), so the graph is concave up. 4. On a concave-up interval, \(f''(x)>0\). Since \(f''\) is the derivative of \(f'\), this means \(f'\) is strictly increasing there; the tangent slopes increase as \(x\) increases.

a) Concave down on \((-\infty, 2)\); concave up on \((2, \infty)\) b) On a concave-up interval, \(f''>0\), so \(f'\) is increasing.

- Find the first two derivatives. - Recall the range of an exponential function. - A strictly positive first derivative cannot have a zero. - An inflection point requires a change in concavity.

1. Differentiate: \(f'(x)=e^{2x}+4\). Since \(e^{2x}>0\), \(f'(x)>0\) for every real \(x\). Therefore, the first derivative is never zero, and the graph has no local extrema. 2. Differentiate again: \(f''(x)=2e^{2x}>0\) for every real \(x\). The concavity never changes, so the graph has no inflection points.

\(f'(x)=e^{2x}+4>0\) and \(f''(x)=2e^{2x}>0\) for every real \(x\). Therefore, the graph has neither local extrema nor inflection points.

- Rewrite the reciprocal functions using negative exponents. - Examine the sign of each denominator for negative and positive \(x\). - How do odd and even powers affect the sign of an expression?

1. For \(g\), \(g'(x)=-x^{-2}\) and \(g''(x)=2x^{-3}=\frac{2}{x^3}\). For \(h\), \(h'(x)=-2x^{-3}\) and \(h''(x)=6x^{-4}=\frac{6}{x^4}\). Both numerators are nonzero constants, so neither second derivative has a zero. 2. For \(g\), \(g''(x)<0\) when \(x<0\) and \(g''(x)>0\) when \(x>0\). Thus, the left branch is concave down and the right branch is concave up. The concavity differs across the gap, but \(x=0\) is not an inflection point because \(g\) is undefined there. 3. For \(h\), \(x^4>0\) whenever \(x\ne0\), so \(h''(x)>0\) on both branches. The graph is concave up throughout its domain.

a) \(g''(x)=\frac{2}{x^3}\ne0\) and \(h''(x)=\frac{6}{x^4}\ne0\) for \(x\ne0\). b) For \(g\), the left branch is concave down and the right branch is concave up; \(x=0\) is not an inflection point. For \(h\), both branches are concave up.

- Relate concavity to the sign of the second derivative. - Determine the sign of \(1-x^2\) on the given interval. - The denominator is positive.

1. Concave up means \(f''(x)>0\) throughout \((-1,1)\). 2. On \((-1,1)\), \(1-x^2>0\), and \((x^2+1)^2>0\). Therefore, expression I is positive throughout the interval. 3. Expression II is the negative of expression I, so it is negative throughout \((-1,1)\). Therefore, only expression I is consistent with the stated concavity.

Expression I: \(f''(x)=\frac{2-2x^2}{(x^2+1)^2}\).

- Relate concavity to the sign of the second derivative. - Determine the sign of \(x^2-1\) on the given interval. - The exponential factor is always positive.

1. Concave down means \(f''(x)<0\) throughout \((-1,1)\). 2. On \((-1,1)\), \(x^2-1<0\), while \(e^{-x^2/2}>0\). Therefore, expression II is negative throughout the interval. 3. Expression I is the negative of expression II, so it is positive throughout \((-1,1)\). Therefore, expression II is consistent with the stated concavity.

Expression II: \(f''(x)=(x^2-1)e^{-x^2/2}\).

- What does “the derivative of \(f'\)” mean? - How are zeros related to possible sign changes? - Can a continuous function change sign where its value is nonzero? - Check the second derivative immediately to the left and right of each value.

1. Differentiate twice: \(f'(x)=2x^3-6x\) and \(f''(x)=6x^2-6=6(x-1)(x+1)\). 2. At \(x=0\), \(f''(0)=-6\ne0\). Since \(f''\) is continuous, it keeps the same sign in a neighborhood of \(0\), so there is no sign change there. 3. At \(x=1\), \(f''(1)=0\). The factor \(x-1\) has odd multiplicity while \(x+1\) remains positive near \(1\), so \(f''\) changes from negative to positive at \(x=1\).

At \(x=0\), the second derivative does not change sign. At \(x=1\), the second derivative changes from negative to positive.

- What must be true of the slope and concavity at a stationary inflection point? - Which derivatives locate a horizontal tangent and a change in concavity? - How does the third derivative provide a sufficient condition? - Why is the point not a local extremum?

1. Differentiate: \(f'(x)=-3x^2+6x-3=-3(x-1)^2\), \(f''(x)=-6x+6\), and \(f'''(x)=-6\). 2. A stationary inflection point requires \(f'(x)=0\) and \(f''(x)=0\). Solving \(-3(x-1)^2=0\) gives \(x=1\), and \(f''(1)=0\). 3. Since \(f'''(1)=-6\ne0\), the graph changes concavity at \(x=1\). Together with \(f'(1)=0\), this proves that the point is a stationary inflection point. 4. Evaluate the function: \(f(1)=-1+3-3+2=1\).

\((1, 1)\)

- Is the slope of a line constant or changing? - What is the second derivative of a linear function? - Which strict signs identify concave up and concave down in this problem? - Is zero positive or negative?

1. For \(f(x)=mx+b\), \(f'(x)=m\) and \(f''(x)=0\). 2. The strict test identifies concave-up behavior only where \(f''(x)>0\) and concave-down behavior only where \(f''(x)<0\). 3. Since \(f''(x)=0\) everywhere, neither strict inequality holds. Under this convention, a line is neither concave up nor concave down and has zero curvature everywhere.

Using the strict sign convention, the claim is correct: a line has \(f''(x)=0\) everywhere, so it is neither concave up nor concave down.

- Find the inflection-point x-coordinate from the second derivative. - Express the parameter in terms of \(x\). - Substitute into the y-coordinate.

1. The derivatives are \(g_t'(x)=3x^2-6tx\), \(g_t''(x)=6x-6t\), and \(g_t'''(x)=6\). 2. The inflection point satisfies \(6x-6t=0\), so \(x=t\). The nonzero third derivative confirms an inflection point. 3. Substitute \(t=x\) into the original function: \(y=x^3-3x\cdot x^2+1=-2x^3+1\).

\(y=-2x^3+1\)

- Look for where the graph changes how it bends. - An inflection point occurs where the concavity changes. - Read the approximate coordinates of each change from the grid.

1. On the displayed interval \([-3.5, 3.5]\), the graph changes concavity near \(x=-1\) and \(x=1\). 2. It is concave up on \([-3.5, -1)\) and \((1, 3.5]\), and concave down on \((-1, 1)\). 3. Reading from the graph gives \(W_1\approx(-1, 0.6)\) and \(W_2\approx(1, 2.6)\).

a) Concave up on \([-3.5, -1)\) and \((1, 3.5]\); concave down on \((-1, 1)\). b) \(W_1\approx(-1, 0.6)\) and \(W_2\approx(1, 2.6)\)

- Identify where the graph changes how it bends. - An inflection point lies where the concavity changes.

1. The graph changes concavity at \(x=2\). 2. On the displayed interval, the graph is concave up on \([-1, 2)\) and concave down on \((2, 6]\). 3. Reading from the graph gives the inflection point \(W\approx(2, 2.6)\).

a) Concave up on \([-1, 2)\); concave down on \((2, 6]\). b) \(W\approx(2, 2.6)\)

- Locate the graph's peaks and valleys. - Look for where the bending direction changes. - Count the x-axis crossings. - Use y-axis symmetry to check paired values.

1. The graph has three local extrema: local minima near \(x=-2.2\) and \(x=2.2\), and a local maximum at \(x=0\). 2. The graph changes concavity near \(x=-1.3\) and \(x=1.3\). Thus, the left inflection point lies between \(-2\) and \(-1\), and the right one lies between \(1\) and \(2\). 3. The graph crosses the x-axis four times on \([-4, 4]\), so four zeros are visible.

a) Three local extrema: two local minima and one local maximum b) Approximately \(x=\pm1.3\); between \(-2\) and \(-1\), and between \(1\) and \(2\) c) Four zeros

- Where does the graph change how it bends? - Set the second derivative equal to zero. - Check the sign of the second derivative on each side of the candidate value.

1. Differentiate twice: \(g'(x)=\frac{1}{2}x^2-2x\) and \(g''(x)=x-2\). 2. Solving \(g''(x)=0\) gives \(x=2\). 3. For \(x<2\), \(g''(x)<0\), so the graph is concave down. For \(x>2\), \(g''(x)>0\), so the graph is concave up. The concavity changes at \(x=2\).

a) The graph changes from concave down to concave up at about \(x=2\). b) Concave down for \(x<2\); concave up for \(x>2\).

- Find where the graph changes from bending upward to bending downward. - Determine whether the tangent slopes are increasing or decreasing on each side.

1. On the displayed domain \([-3, 5]\), the graph changes concavity at \(x=1\). 2. The graph is concave up on \([-3, 1)\) and concave down on \((1, 5]\).

Concave up on \([-3, 1)\); concave down on \((1, 5]\).

- Locate the points where the graph changes concavity. - Those values divide the displayed domain into concavity intervals. - Classify the graph’s bending on each open interval.

1. The graph changes concavity at \(x=-2\) and \(x=2\). 2. On the displayed domain \([-6, 6]\), the graph is concave up on \([-6, -2)\) and \((2, 6]\), and concave down on \((-2, 2)\).

Concave up on \([-6, -2)\) and \((2, 6]\); concave down on \((-2, 2)\).

- Determine the sign of \(\sin(x)\) on each side of \(0\). - Use the \(\pi\)-based labels on the x-axis.

1. The second derivative is \(f''(x)=-2\sin(x)\). 2. On \([-\pi, 0)\), \(\sin(x)<0\), so \(f''(x)>0\) and the graph is concave up. 3. On \((0, \pi]\), \(\sin(x)>0\), so \(f''(x)<0\) and the graph is concave down. 4. The concavity changes at \(x=0\).

Concave up on \([-\pi, 0)\); concave down on \((0, \pi]\).

- Are the negative slopes becoming steeper or less steep? - How does concavity determine the sign of the second derivative?

1. The temperature is decreasing, but the graph becomes less steep as time passes. 2. Therefore, the slope is negative but increasing toward \(0\). The graph is concave up. 3. A concave-up graph has a positive second derivative, so \(T''(t)>0\) throughout the displayed interval.

\(T''(t)>0\) throughout the displayed interval because the graph is concave up.

- Find where the graph changes concavity. - What sign does the second derivative have on a concave-down or concave-up interval?

1. The graph changes from concave down to concave up at about \(x=2\). The point is approximately \(W=\left(2, \frac{4}{3}\right)\). 2. For \(x<2\), the graph is concave down, so \(f''(x)<0\). 3. For \(x>2\), the graph is concave up, so \(f''(x)>0\). 4. Thus the second derivative changes from negative to positive at \(x=2\).

The inflection value is approximately \(x=2\). The second derivative changes from negative for \(x<2\) to positive for \(x>2\).

- Look for where the graph changes between cup-shaped and cap-shaped bending. - Use the graph’s symmetry to refine your estimates.

1. From the graph, the concavity changes near \(x=-1.4\) and \(x=1.4\). 2. On the displayed interval \([-5, 5]\), the graph is concave up on \((-1.4, 1.4)\). 3. It is concave down on \([-5, -1.4)\) and \((1.4, 5]\).

Inflection values: approximately \(x=-1.4\) and \(x=1.4\). Concave up on \((-1.4, 1.4)\); concave down on \([-5, -1.4)\) and \((1.4, 5]\).

- Differentiate the relationship \(F^{\prime}=f\). - Inflection points occur where the second derivative changes sign. - Look for local extrema of \(f\).

1. Since \(F^{\prime}=f\), the second derivative of \(F\) is \(F^{\prime\prime}=f^{\prime}\). 2. An inflection point of \(F\) occurs where \(F^{\prime\prime}=f^{\prime}\) changes sign. This happens at local extrema of \(f\). 3. From the graph, \(f\) has a local minimum at approximately \(x=-2.3\) and a local maximum at approximately \(x=2.3\). Therefore, every antiderivative \(F\) has inflection points at those x-values.

\(x\approx-2.3\) and \(x\approx2.3\)

- Differentiate the relationship \(F^{\prime}=f\). - Concavity up corresponds to a positive second derivative. - Find where \(f\) is increasing.

1. Since \(F^{\prime}=f\), \(F^{\prime\prime}=f^{\prime}\). 2. The graph of \(F\) is concave up where \(F^{\prime\prime}>0\), which is where \(f^{\prime}>0\), or equivalently where \(f\) is increasing. 3. The graph of \(f\) is increasing from its local minimum at \(x=0\) to its local maximum at \(x=2\). Therefore, every antiderivative \(F\) is concave up on \((0, 2)\).

\((0, 2)\)

- Which derivative describes concavity and helps locate inflection points? - What equation must that derivative satisfy at \(x = 2\)? - Rearrange the equation so the parameter is alone.

1. Differentiate twice: \(y' = 3x^2 - 2ax\) and \(y'' = 6x - 2a\). 2. At an inflection point, the second derivative must be zero, so \(y''(2) = 0\). 3. Solve \(6 \cdot 2 - 2a = 0\): \(12 - 2a = 0\), so \(a = 6\). Because \(y''(x) = 6x - 12\) changes sign at \(x = 2\), the point is an inflection point.

\(a = 6\)

- Distinguish necessary conditions from sufficient conditions. - Look for a function whose second derivative is \(0\) at a point but does not change sign. - A stationary inflection point must have both a horizontal tangent and a change in concavity. - Consider a function with a flat local minimum.

1. If the concavity changes at \(x_0\) and \(f''\) is continuous, then \(f''\) changes sign and must equal \(0\) at \(x_0\). Thus \(f''(x_0)=0\) is necessary. 2. The condition is not sufficient. For \(f(x)=x^4\), \(f''(x)=12x^2\), so \(f''(0)=0\), but \(f''\) is positive on both sides of \(0\). The graph has a local minimum, not an inflection point. 3. The conditions in part b) are also not sufficient. For the same function, \(f'(0)=0\) and \(f''(0)=0\), but there is no change in concavity and therefore no stationary inflection point.

a) The condition is necessary by continuity of \(f''\), but it is not sufficient. The function \(f(x)=x^4\) has \(f''(0)=0\) without an inflection point. b) No. For \(f(x)=x^4\), both \(f'(0)=0\) and \(f''(0)=0\), yet \(x=0\) is a local minimum, not a stationary inflection point.

- Think about how the tangent slope, represented by \(f'\), changes when a graph is concave up or concave down. - Recall the difference between a necessary condition and a sufficient condition. - Can the second derivative equal zero without changing sign? - What does monotonic behavior of \(f'\) tell you about the shape of \(f\)?

1. Statement a is true. A positive second derivative means the slope function \(f'\) is increasing, so \(f\) is concave up. 2. Statement b is true. Because \(f''\) is continuous, an inflection point requires a change in the sign of \(f''\), and continuity then gives \(f''(x_0)=0\). 3. Statement c is false. For example, if \(f(x)=x^4\), then \(f''(0)=0\), but \(f''(x)=12x^2\) does not change sign at \(x=0\), so there is no inflection point. 4. Statement d is true. A strictly decreasing derivative means the tangent slopes decrease as \(x\) increases, which is the defining behavior of a concave-down function.

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

- Determine where the second derivative is positive and where it is negative. - What sign change is required for an inflection point? - Combine the concavity information with the slope at \(x=4\). - What do you call an inflection point with a horizontal tangent?

1. Solve \(f''(x)=0\): \(0.5x-2=0\), so \(x=4\). 2. For \(x<4\), \(f''(x)<0\), so \(f\) is concave down. For \(x>4\), \(f''(x)>0\), so \(f\) is concave up. 3. Since \(f''\) changes sign from negative to positive at \(x=4\), the graph has an inflection point there. 4. If \(f'(4)=0\), the tangent at the inflection point is horizontal. Therefore, the point is a stationary inflection point.

a) Concave down on \((-\infty, 4)\) and concave up on \((4, \infty)\) b) \(f''(4)=0\), and \(f''\) changes sign at \(x=4\), so there is an inflection point. c) A stationary inflection point, because the inflection point has a horizontal tangent.

- What does the sign of the second derivative tell you about concavity? - For \(f'\) to have a local extremum, what must its derivative do? - How does the second derivative test classify a critical point of \(f\)? - Use the factored form to make a sign chart.

1. The zeros of \(f''\) are \(-4\), \(-1\), and \(2\). A sign chart gives \(f''(x)<0\) on \((-\infty, -4)\) and \((-1, 2)\), so \(f\) is concave down on those intervals. 2. Local extrema of \(f'\) occur where its derivative \(f''\) changes sign. Each zero of \(f''\) has odd multiplicity, so \(f''\) changes sign at \(x=-4\), \(x=-1\), and \(x=2\). Thus, \(f'\) has a local extremum at each of those values. 3. Since \(f'(-5)=0\) and \(f''(-5)=(-1)(-4)(-7)=-28<0\), the second derivative test gives a local maximum at \(x=-5\). Since \(f'(3)=0\) and \(f''(3)=7 \cdot 4 \cdot 1=28>0\), the second derivative test gives a local minimum at \(x=3\).

a) \((-\infty, -4)\) and \((-1, 2)\) b) \(x=-4\), \(x=-1\), and \(x=2\) c) A local maximum at \(x=-5\) and a local minimum at \(x=3\)

- What two facts establish an inflection point? - On which intervals is the second derivative positive? - First factor \(x^2-16\) to locate all possible sign changes. - Use the sign of the second derivative at each critical point.

1. At \(x=0\), \(f''(0)=0\). Also, \(f''(-1)=-7.5<0\) and \(f''(1)=7.5>0\), so \(f''\) changes sign at \(x=0\). Therefore, the graph has an inflection point there. 2. The zeros of \(f''\) are \(-4\), \(0\), and \(4\). A sign chart shows that \(f''(x)>0\) on \((-\infty, -4)\) and \((0, 4)\). Therefore, the graph is concave up on those intervals. 3. Since \(f'(-2)=0\) and \(f''(-2)=-12<0\), \(f\) has a local maximum at \(x=-2\). Since \(f'(2)=0\) and \(f''(2)=12>0\), \(f\) has a local minimum at \(x=2\).

a) \(f''(0)=0\), and \(f''\) changes sign from negative to positive at \(x=0\). b) \((-\infty, -4)\) and \((0, 4)\) c) A local maximum at \(x=-2\) and a local minimum at \(x=2\)

- First find the second derivative. - When does an equation of the form \(ax^2+c=0\) have zero, one, or two real solutions? - A zero of the second derivative is an inflection point only when the second derivative changes sign. - How does the vertical position of the parabola \(y=f_k''(x)\) depend on \(k\)?

1. Differentiate twice: \(f_k'(x)=4x^3+2kx\) and \(f_k''(x)=12x^2+2k\). 2. Possible inflection points satisfy \(12x^2+2k=0\), so \(x^2=-\frac{k}{6}\). 3. If \(k>0\), the equation has no real solution and \(f_k''(x)>0\) for every \(x\). The graph is concave up everywhere and has no inflection points. 4. If \(k=0\), then \(f_0''(x)=12x^2\). It equals zero at \(x=0\) but does not change sign, so the graph has no inflection point. 5. If \(k<0\), the second derivative has two zeros, \(x=\pm\sqrt{-\frac{k}{6}}\). Because \(f_k''\) is an upward-opening quadratic, it changes sign at both zeros, so the graph has two inflection points.

a) \(f_k''(x)=12x^2+2k\) b) The graph does not change concavity when \(k\ge0\). c) There are no inflection points when \(k\ge0\), and there are two inflection points when \(k<0\).

- Which derivative determines concavity? - How does the sign of the second derivative distinguish concave up from concave down? - What sign change must occur at an inflection point? - Substitute each inflection value into the original function to find its \(y\)-coordinate.

1. Differentiate twice: \(f'(x)=\frac{1}{3}x^3-x+2\) and \(f''(x)=x^2-1\). 2. Solve \(f''(x)=0\): \(x^2-1=0\), so \(x=-1\) and \(x=1\). 3. Since \(x^2-1>0\) for \(x<-1\) and \(x>1\), the graph is concave up on those intervals. Since \(x^2-1<0\) for \(-1<x<1\), the graph is concave down there. 4. Evaluate the function at the inflection values: \(f(-1)=\frac{1}{12}-\frac{1}{2}-2=-\frac{29}{12}\) and \(f(1)=\frac{1}{12}-\frac{1}{2}+2=\frac{19}{12}\).

1. Concave up on \((-\infty, -1)\cup(1, \infty)\); concave down on \((-1, 1)\) 2. \(\left(-1, -\frac{29}{12}\right)\) and \(\left(1, \frac{19}{12}\right)\)

- Use the second derivative. - What can you always say about \(x^2\)? - How does the sign of \(t\) affect the sign of the second derivative? - Remember to reverse an inequality when dividing by a negative number.

1. Differentiate twice: \(g_t'(x)=4tx^3+12x\) and \(g_t''(x)=12tx^2+12=12(tx^2+1)\). 2. If \(t\ge0\), then \(tx^2\ge0\) for every real \(x\). Thus, \(g_t''(x)=12(tx^2+1)\ge12>0\), so the graph is concave up everywhere. 3. If \(t<0\), the graph is concave down when \(12(tx^2+1)<0\). This is equivalent to \(tx^2<-1\). 4. Dividing by the negative number \(t\) reverses the inequality: \(x^2>-\frac{1}{t}\). Therefore, \(x<-\sqrt{-\frac{1}{t}}\) or \(x>\sqrt{-\frac{1}{t}}\).

1. For \(t\ge0\), \(g_t''(x)=12(tx^2+1)\ge12>0\), so the graph is concave up for all real \(x\). 2. For \(t<0\), the graph is concave down when \(x<-\sqrt{-\frac{1}{t}}\) or \(x>\sqrt{-\frac{1}{t}}\).

- Which derivative determines concavity? - What does the sign of the second derivative tell you? - At which values can a graph change concavity? - Use the first derivative to find critical points and the second derivative to classify them.

1. Differentiate: \(f_k'(x)=3x^2-6kx\), \(f_k''(x)=6x-6k\), and \(f_k'''(x)=6\). 2. Solve \(f_k''(x)=0\): \(6x-6k=0\), so \(x=k\). The second derivative is linear with nonzero slope, so this is its only zero and it changes sign there. 3. Since \(f_k''(x)=6(x-k)\), it is negative for \(x<k\) and positive for \(x>k\). Thus, the graph changes from concave down to concave up. 4. Solve \(f_k'(x)=3x(x-2k)=0\): the critical values are \(x=0\) and \(x=2k\). 5. Since \(f_k''(0)=-6k<0\), \(f_k\) has a local maximum at \((0, 2)\). Since \(f_k''(2k)=6k>0\), \(f_k\) has a local minimum at \((2k, 2-4k^3)\).

a) The only inflection value is \(x=k\). b) The graph is concave down for \(x<k\) and concave up for \(x>k\). c) Local maximum: \((0, 2)\). Local minimum: \((2k, 2-4k^3)\).

- Use the necessary and sufficient conditions for an inflection point. - How does the sign of the second derivative determine concavity? - To classify extrema, combine zeros of the first derivative with the sign of the second derivative.

1. Differentiate: \(h_a'(x)=ax^2-4x\), \(h_a''(x)=2ax-4\), and \(h_a'''(x)=2a\). 2. Solve \(h_a''(x)=0\): \(2ax-4=0\), so \(x=\frac{2}{a}\). Since \(a>0\), \(h_a'''(x)=2a\ne0\), so the second derivative changes sign there. 3. If \(x<\frac{2}{a}\), then \(2ax-4<0\), so the graph is concave down. If \(x>\frac{2}{a}\), then \(2ax-4>0\), so the graph is concave up. 4. At \(x=0\), \(h_a'(0)=0\) and \(h_a''(0)=-4<0\), so \(h_a\) has a local maximum there. 5. Factoring \(h_a'(x)=x(ax-4)\) gives the other critical value \(x=\frac{4}{a}\). Since \(h_a''\left(\frac{4}{a}\right)=4>0\), this critical point is a local minimum.

a) \(x=\frac{2}{a}\) b) Concave down for \(x<\frac{2}{a}\); concave up for \(x>\frac{2}{a}\) c) Local maximum at \(x=0\); local minimum at \(x=\frac{4}{a}\)

- Expanding the expression first may make differentiation easier. - What equation must the second derivative satisfy at a possible inflection point? - How many solutions can the resulting quadratic equation have? - Confirm that the second derivative changes sign at each candidate.

1. Expand the function: \(g(x)=\frac{1}{4}x^4-\frac{3}{2}x^3+3x^2\). 2. Differentiate: \(g'(x)=x^3-\frac{9}{2}x^2+6x\), \(g''(x)=3x^2-9x+6\), and \(g'''(x)=6x-9\). 3. Solve \(g''(x)=0\): \(3(x^2-3x+2)=3(x-1)(x-2)=0\), so \(x=1\) and \(x=2\). 4. Since \(g'''(1)=-3\ne0\) and \(g'''(2)=3\ne0\), the graph changes concavity at both values.

\(x=1\) and \(x=2\)

- What conditions establish an inflection point? - Which derivative gives the slope of a tangent line? - What extra condition distinguishes a stationary inflection point? - Check whether the first derivative is zero at the inflection value.

1. Differentiate: \(f'(x)=\frac{1}{4}x^2-2x+3\), \(f''(x)=\frac{1}{2}x-2\), and \(f'''(x)=\frac{1}{2}\). 2. Solve \(f''(x)=0\): \(\frac{1}{2}x-2=0\), so \(x=4\). Since \(f'''(4)=\frac{1}{2}\ne0\), the graph changes concavity there. 3. Evaluate the function: \(f(4)=\frac{1}{12}\cdot4^3-4^2+3\cdot4-1=\frac{1}{3}\). The inflection point is \(\left(4, \frac{1}{3}\right)\). 4. The tangent slope is \(f'(4)=\frac{1}{4}\cdot4^2-2\cdot4+3=-1\). 5. Because \(f'(4)=-1\ne0\), the tangent is not horizontal, so the point is not a stationary inflection point.

a) \(\left(4, \frac{1}{3}\right)\) b) \(-1\) c) No. The tangent is not horizontal because \(f'(4)=-1\).

- A function can have more than one inflection point. - Which derivative gives the tangent slope? - What must be true of the first derivative at a stationary inflection point? - Substitute each inflection value into the original function to find the point coordinates.

1. Differentiate: \(f'(x)=x^3-3x^2\), \(f''(x)=3x^2-6x\), and \(f'''(x)=6x-6\). 2. Solve \(f''(x)=3x(x-2)=0\): \(x=0\) and \(x=2\). Since \(f'''(0)=-6\ne0\) and \(f'''(2)=6\ne0\), both values give inflection points. 3. Evaluate the function: \(f(0)=2\) and \(f(2)=\frac{1}{4}\cdot16-8+2=-2\). The points are \((0, 2)\) and \((2, -2)\). 4. Evaluate the first derivative: \(f'(0)=0\) and \(f'(2)=8-12=-4\). 5. The point \((0, 2)\) is a stationary inflection point because its tangent is horizontal. The point \((2, -2)\) is not stationary because its tangent slope is \(-4\).

a) \((0, 2)\) and \((2, -2)\) b) At \((0, 2)\), the slope is \(0\). At \((2, -2)\), the slope is \(-4\). c) \((0, 2)\) is a stationary inflection point; \((2, -2)\) is not.

- Where can a graph change concavity? - Relate the sign of the second derivative to concavity. - What does concave up imply about the tangent slope as \(x\) increases? - Test one value between the two zeros of the second derivative.

1. Differentiate: \(g'(x)=-\frac{1}{3}x^3+2x\) and \(g''(x)=-x^2+2\). 2. Solve \(g''(x)=0\): \(-x^2+2=0\), so \(x=-\sqrt{2}\) and \(x=\sqrt{2}\). 3. The graph of \(y=-x^2+2\) is above the \(x\)-axis between its zeros. Therefore, \(g''(x)>0\) for \(-\sqrt{2}<x<\sqrt{2}\), so \(g\) is concave up there. 4. Since \(g''>0\) on this interval, the derivative \(g'\) is strictly increasing. Thus, the tangent slope increases as \(x\) moves from left to right.

a) \(x=-\sqrt{2}\) and \(x=\sqrt{2}\) b) Since \(g''(x)=-x^2+2>0\) on \((-\sqrt{2}, \sqrt{2})\), the graph is concave up and \(g'\) is increasing there.

- Which derivatives are needed to locate and verify inflection points? - What equation gives possible inflection values? - Use point-slope form to write each tangent line. - What slope identifies a stationary inflection point?

1. Differentiate: \(f'(x)=x^3-3x\), \(f''(x)=3x^2-3\), and \(f'''(x)=6x\). 2. Solve \(f''(x)=0\): \(3x^2-3=0\), so \(x=-1\) and \(x=1\). Since \(f'''(-1)=-6\ne0\) and \(f'''(1)=6\ne0\), both values give inflection points. 3. Evaluate the function: \(f(-1)=f(1)=-\frac{5}{4}\). Thus, the inflection points are \(\left(-1, -\frac{5}{4}\right)\) and \(\left(1, -\frac{5}{4}\right)\). 4. The tangent slopes are \(f'(-1)=2\) and \(f'(1)=-2\). Using point-slope form gives \(y=2(x+1)-\frac{5}{4}=2x+\frac{3}{4}\) and \(y=-2(x-1)-\frac{5}{4}=-2x+\frac{3}{4}\). 5. Neither inflection point is stationary because neither tangent slope is zero.

Inflection points: \(\left(-1, -\frac{5}{4}\right)\) and \(\left(1, -\frac{5}{4}\right)\) Tangent lines: \(y=2x+\frac{3}{4}\) and \(y=-2x+\frac{3}{4}\) Neither point is a stationary inflection point.

- How many derivatives do you need to analyze concavity and inflection points? - What must the tangent slope be at a stationary inflection point? - Use the point and slope to write each tangent line.

1. Differentiate: \(g'(x)=4x^3-12x^2\), \(g''(x)=12x^2-24x\), and \(g'''(x)=24x-24\). 2. Solve \(g''(x)=12x(x-2)=0\): \(x=0\) and \(x=2\). Since \(g'''(0)=-24\ne0\) and \(g'''(2)=24\ne0\), both are inflection values. 3. At \(x=0\), \(g(0)=0\) and \(g'(0)=0\). The tangent line is \(y=0\), and \((0, 0)\) is a stationary inflection point. 4. At \(x=2\), \(g(2)=-16\) and \(g'(2)=-16\). The tangent line is \(y+16=-16(x-2)\), or \(y=-16x+16\). Because the slope is not zero, \((2, -16)\) is not stationary.

Inflection points: \((0, 0)\) and \((2, -16)\) Tangent lines: \(y=0\) and \(y=-16x+16\) The point \((0, 0)\) is a stationary inflection point; \((2, -16)\) is not.

- Use the first and second derivatives to locate and classify extrema. - Substitute an x-coordinate into the original function to find the corresponding y-coordinate. - The tangent slope equals the value of the first derivative at the point of tangency. - Use point-slope form for the tangent line.

1. Differentiate: \(f'(x)=x^2-2x-3\), \(f''(x)=2x-2\), and \(f'''(x)=2\). 2. Solve \(f'(x)=0\): \((x+1)(x-3)=0\), so the critical values are \(x=-1\) and \(x=3\). Since \(f''(-1)=-4<0\), \((-1, \frac{11}{3})\) is a local maximum. Since \(f''(3)=4>0\), \((3, -7)\) is a local minimum. 3. Solve \(f''(x)=0\): \(2x-2=0\), so \(x=1\). Because \(f'''(1)=2\ne0\), the graph has an inflection point. Since \(f(1)=-\frac{5}{3}\), the inflection point is \((1, -\frac{5}{3})\). Its slope is \(f'(1)=-4\). 4. Using point-slope form, \(y+\frac{5}{3}=-4(x-1)\), so the tangent line is \(y=-4x+\frac{7}{3}\).

a) Local maximum: \((-1, \frac{11}{3})\); local minimum: \((3, -7)\) b) Inflection point: \((1, -\frac{5}{3})\); slope: \(-4\) c) \(y=-4x+\frac{7}{3}\)

- Use the necessary and sufficient derivative tests for extrema and inflection points. - A tangent line requires both the point of tangency and the slope there. - To find an axis intercept, set the other coordinate equal to zero.

1. Differentiate: \(g'(x)=-3x^2+12x-9\), \(g''(x)=-6x+12\), and \(g'''(x)=-6\). 2. Solve \(g'(x)=0\): \(-3(x-1)(x-3)=0\), so \(x=1\) and \(x=3\). Since \(g''(1)=6>0\), \((1, 0)\) is a local minimum. Since \(g''(3)=-6<0\), \((3, 4)\) is a local maximum. 3. Solve \(g''(x)=0\): \(x=2\). Since \(g'''(2)=-6\ne0\), the graph has an inflection point at \((2, 2)\). 4. The slope at the inflection point is \(g'(2)=3\). Point-slope form gives \(y-2=3(x-2)\), or \(y=3x-4\). 5. Setting \(x=0\) gives the y-intercept \((0, -4)\). Setting \(y=0\) gives \(x=\frac{4}{3}\), so the x-intercept is \((\frac{4}{3}, 0)\).

a) Local minimum: \((1, 0)\); local maximum: \((3, 4)\); inflection point: \((2, 2)\) b) \(y=3x-4\) c) x-intercept: \((\frac{4}{3}, 0)\); y-intercept: \((0, -4)\)

- Does a vertical shift change the \(x\)-coordinate of a point? - How does multiplying the entire output of a function affect its second derivative? - What input to \(f\) corresponds to the original inflection value? - Relate a factor inside the function to a horizontal scaling.

1. The function \(p(x)=f(x)+8\) shifts the graph up \(8\) units. A vertical shift does not change the \(x\)-coordinate of an inflection point, so the inflection value remains \(x=3\). 2. The function \(q(x)=-2f(x)\) reflects the graph across the \(x\)-axis and stretches it vertically by a factor of \(2\). These vertical transformations do not change the inflection value, so it remains \(x=3\). 3. The function \(r(x)=f(3x)\) compresses the graph horizontally by a factor of \(\frac{1}{3}\). The original input \(3\) occurs when \(3x=3\), so the new inflection value is \(x=1\).

a) \(x=3\); shift up \(8\) units b) \(x=3\); reflection across the \(x\)-axis and vertical stretch by a factor of \(2\) c) \(x=1\); horizontal compression by a factor of \(\frac{1}{3}\)

- Which derivatives are needed to locate an inflection point? - Set the second derivative equal to zero. - Treat the parameter \(t\) as a constant while differentiating with respect to \(x\). - Substitute the inflection value into the original function.

1. Differentiate: \(g_t'(x)=\frac{1}{2}x^2-2tx+10\), \(g_t''(x)=x-2t\), and \(g_t'''(x)=1\). 2. Solve \(g_t''(x)=0\): \(x-2t=0\), so the inflection value is \(x=2t\). Since \(g_t'''(x)=1\ne0\), the graph changes concavity there. 3. Evaluate the function: \(g_t(2t)=\frac{1}{6}(2t)^3-t(2t)^2+10(2t)=\frac{4}{3}t^3-4t^3+20t=-\frac{8}{3}t^3+20t\).

\(W_t=\left(2t, -\frac{8}{3}t^3+20t\right)\)

- Find the first and second derivatives with respect to \(x\). - Set the second derivative equal to zero and solve for \(x\) in terms of \(k\). - Use exponent rules carefully when substituting a fraction into the function. - Why is the restriction \(k\ne0\) necessary?

1. Differentiate: \(h_k'(x)=3kx^2+18x\), \(h_k''(x)=6kx+18\), and \(h_k'''(x)=6k\). 2. Solve \(h_k''(x)=0\): \(6kx+18=0\), so \(x=-\frac{3}{k}\). 3. Since \(k\ne0\), \(h_k'''(x)=6k\ne0\), so the graph changes concavity at this value. 4. Evaluate the function: \(h_k\left(-\frac{3}{k}\right)=k\left(-\frac{3}{k}\right)^3+9\left(-\frac{3}{k}\right)^2-5=-\frac{27}{k^2}+\frac{81}{k^2}-5=\frac{54}{k^2}-5\).

\(W_k=\left(-\frac{3}{k}, \frac{54}{k^2}-5\right)\)

- What must the second derivative equal at a possible inflection point? - Differentiate twice with respect to \(x\), treating \(a\) as a constant. - Substitute \(x=5\) into the second derivative and solve for \(a\).

1. Differentiate twice: \(f_a'(x)=x^2+2(a-3)x+7\) and \(f_a''(x)=2x+2a-6\). 2. Require \(f_a''(5)=0\): \(2\cdot5+2a-6=0\), so \(4+2a=0\). 3. Solving gives \(a=-2\). 4. For this parameter, \(f_{-2}'''(x)=2\ne0\), so the graph does change concavity at \(x=5\).

\(a=-2\)

- What is the \(y\)-coordinate of every point on the \(x\)-axis? - First find the inflection value using the second derivative. - Substitute that value into the original function. - Set the resulting function value equal to zero.

1. Differentiate: \(g_k'(x)=3x^2-6x\) and \(g_k''(x)=6x-6\). 2. Solve \(g_k''(x)=0\): \(6x-6=0\), so the inflection value is \(x=1\). Since \(g_k'''(x)=6\ne0\), the graph changes concavity there. 3. A point on the \(x\)-axis has \(y=0\), so require \(g_k(1)=0\). Then \(1-3+k=0\), which gives \(k=2\).

\(k=2\)

- Consider a fourth-degree polynomial shifted upward. - Distinguish a stationary inflection point from a local extremum. - Determine the degree of the second derivative of a cubic.

1. Statement a) is false. The polynomial \(f(x)=x^4+1\) satisfies \(f(x)\geq1\), so it has no real zeros. 2. Statement b) is false. For \(f(x)=x^3\), \(f'(x)=3x^2\) is zero only at \(x=0\) and does not change sign there, so the function has no local extremum. 3. Statement c) is true. A cubic has the form \(f(x)=ax^3+bx^2+cx+d\), with \(a\neq0\). Its second derivative is \(f''(x)=6ax+2b\), which has exactly one zero and changes sign there. Thus the cubic has exactly one inflection point.

a) False; for example, \(f(x)=x^4+1\) b) False; for example, \(f(x)=x^3\) c) True

- Examine the possible zeros of the first and second derivatives. - A local extremum requires a sign change in the first derivative. - Simple power functions can provide useful counterexamples. - Determine the degree of the second derivative of a quartic.

1. Statement a) is false. For example, \(f(x)=x^3+x\) has \(f'(x)=3x^2+1>0\) for every real \(x\), so it has no local extrema. 2. Statement b) is true. For \(f(x)=x^4-2x^2\), \(f'(x)=4x(x^2-1)\) has three simple zeros, so its sign changes at \(x=-1\), \(x=0\), and \(x=1\). The function has three local extrema. 3. Statement c) is true. The second derivative of a fourth-degree polynomial is quadratic, so it has at most two real zeros. Therefore, the original polynomial can have at most two inflection points.

a) False b) True c) True

- Find the inflection value in terms of the parameter. - Which derivatives are relevant to concavity and inflection points? - A point lies on a graph when its coordinates satisfy the function equation. - Substitute the inflection value into the function and use the given \(y\)-coordinate.

1. Differentiate: \(f_a'(x)=3x^2-6ax\) and \(f_a''(x)=6x-6a\). 2. Solve \(f_a''(x)=0\): \(6x-6a=0\), so the inflection value is \(x=a\). Since \(f_a'''(x)=6\ne0\), the graph changes concavity there for every \(a\). 3. Require the inflection point to have \(y\)-coordinate \(6\): \(f_a(a)=6\). 4. Compute \(a^3-3a\cdot a^2+4=-2a^3+4=6\). Thus, \(a^3=-1\), so \(a=-1\).

\(a=-1\)

- Distinguish between an inflection point and its \(x\)-coordinate. - First find the two possible inflection values. - How do you find the distance between two values on the \(x\)-axis? - Write an equation involving only \(k\).

1. Differentiate: \(g_k'(x)=\frac{1}{3}x^3-kx+5\) and \(g_k''(x)=x^2-k\). 2. Solve \(g_k''(x)=0\): \(x=\pm\sqrt{k}\). Since \(k>0\), both values are real, and \(g_k'''(x)=2x\ne0\) at each one. 3. The horizontal distance between the inflection points is \(\sqrt{k}-(-\sqrt{k})=2\sqrt{k}\). 4. Set \(2\sqrt{k}=6\). Then \(\sqrt{k}=3\), so \(k=9\).

\(k=9\)

- Use the second derivative to locate the inflection value. - Substitute the point into the line equation. - Determine the horizontal and vertical components of the translation. - A shift right by \(d\) replaces \(x\) with \(x-d\).

1. The derivatives are \(f'(x)=3x^2-6x-9\), \(f''(x)=6x-6\), and \(f'''(x)=6\). Solving \(f''(x)=0\) gives \(x=1\), and \(f'''(1)\ne0\), so this is an inflection value. 2. Since \(f(1)=-6\), the inflection point is \((1, -6)\). Substitution into the line gives \(-4(1)-2=-6\), so the point lies on the line. 3. The point moves \(2\) units right and \(4\) units down. Therefore, \(h(x)=f(x-2)-4\). Thus, \(h(x)=(x-2)^3-3(x-2)^2-9(x-2)+1\), which expands to \(h(x)=x^3-9x^2+15x-1\).

a) The inflection point is \((1, -6)\), and \(-6=-4(1)-2\). b) \(h(x)=(x-2)^3-3(x-2)^2-9(x-2)+1\), or \(h(x)=x^3-9x^2+15x-1\)

- Set the second derivative equal to zero. - Verify the point by substituting into the line equation. - Determine which translation sends the inflection point to the origin. - Expand \((x+2)^3\) carefully and combine like terms.

1. The derivatives are \(g'(x)=\frac{3}{2}x^2-6x+4\) and \(g''(x)=3x-6\). Solving \(g''(x)=0\) gives \(x=2\). Since \(g(2)=0\), the inflection point is \(W=(2, 0)\). 2. The line gives \(-2+2=0\) at \(x=2\), so \(W\) lies on \(y=-x+2\). 3. Moving \((2, 0)\) to the origin translates the graph \(2\) units left. Therefore, \(k(x)=g(x+2)\). Expanding gives \(k(x)=\frac{1}{2}(x+2)^3-3(x+2)^2+4(x+2)=\frac{1}{2}x^3-2x\).

a) \(W=(2, 0)\), and \(0=-2+2\). b) \(k(x)=\frac{1}{2}x^3-2x\)

- How do you locate a change in concavity? - Which derivatives are needed to find and verify an inflection point? - Use the point and tangent slope to write a line equation. - Sketch the intercepts of the tangent line. - The coordinate axes form the perpendicular legs of the triangle.

1. Differentiate: \(f'(x)=\frac{3}{4}x^2-3x\), \(f''(x)=\frac{3}{2}x-3\), and \(f'''(x)=\frac{3}{2}\). 2. Solve \(f''(x)=0\): \(\frac{3}{2}x-3=0\), so \(x=2\). Since \(f'''(2)\ne0\), the graph has an inflection point there. 3. Evaluate the function: \(f(2)=2-6+2=-2\), so \(W=(2, -2)\). 4. The tangent slope is \(f'(2)=3-6=-3\). Thus, \(y+2=-3(x-2)\), or \(y=-3x+4\). 5. The intercepts are \(A=\left(\frac{4}{3}, 0\right)\) and \(B=(0, 4)\). 6. The triangle is right, so its area is \(\frac{1}{2}\cdot\frac{4}{3}\cdot4=\frac{8}{3}\) square units.

a) \(W=(2, -2)\) and \(t: y=-3x+4\) b) \(\frac{8}{3}\) square units

- Set the second derivative equal to zero to find the inflection value. - Use point-slope form or \(y=mx+b\) for the tangent line. - Which derivative gives the tangent slope?

1. Differentiate: \(g'(x)=\frac{1}{2}x^2-2x+3\), \(g''(x)=x-2\), and \(g'''(x)=1\). 2. Solve \(g''(x)=0\): \(x=2\). Since \(g'''(2)=1\ne0\), the graph changes concavity there. 3. Evaluate the function: \(g(2)=\frac{8}{6}-4+6=\frac{10}{3}\), so \(W=\left(2, \frac{10}{3}\right)\). 4. The tangent slope is \(g'(2)=2-4+3=1\). Point-slope form gives \(y-\frac{10}{3}=x-2\), so \(y=x+\frac{4}{3}\).

\(W=\left(2, \frac{10}{3}\right)\); tangent line \(y=x+\frac{4}{3}\)

- Use the second derivative to find the inflection point in terms of the parameter. - Write both coordinates as expressions involving \(a\). - Solve the \(x\)-coordinate equation for \(a\). - Substitute to eliminate the parameter. - Use the restriction \(a>0\) to determine the domain of the locus.

1. Differentiate: \(g_a'(x)=3x^2+6ax\) and \(g_a''(x)=6x+6a\). 2. Solve \(g_a''(x)=0\): \(x=-a\). Since \(g_a'''(x)=6\ne0\), this value gives an inflection point. 3. Its \(y\)-coordinate is \(g_a(-a)=(-a)^3+3a(-a)^2-2a=2a^3-2a\). 4. From \(x=-a\), obtain \(a=-x\). Substitute into the \(y\)-coordinate: \(y=2(-x)^3-2(-x)=-2x^3+2x\). 5. Because \(a>0\), the inflection-point coordinate satisfies \(x=-a<0\).

\(y=-2x^3+2x\) for \(x<0\)

- Which derivative determines concavity? - What do positive and negative second-derivative values mean? - At which values can concavity change? - How can you confirm that a zero of the second derivative is an inflection value?

1. Differentiate: \(h'(x)=2x^3-6x+5\), \(h''(x)=6x^2-6\), and \(h'''(x)=12x\). 2. Solve \(h''(x)=0\): \(6x^2-6=0\), so \(x=-1\) and \(x=1\). 3. The second derivative is positive for \(x<-1\) and \(x>1\), so the graph is concave up there. It is negative for \(-1<x<1\), so the graph is concave down there. 4. Since \(h'''(-1)=-12\ne0\) and \(h'''(1)=12\ne0\), both values give inflection points. 5. Evaluate the function: \(h(-1)=-\frac{19}{2}\) and \(h(1)=\frac{1}{2}\).

a) Concave up on \((-\infty, -1)\cup(1, \infty)\); concave down on \((-1, 1)\) b) \(\left(-1, -\frac{19}{2}\right)\) and \(\left(1, \frac{1}{2}\right)\)

- Which derivatives locate inflection points? - What is the point-slope equation of a tangent line at \(x=x_0\)? - How does the symmetry of the function affect the inflection points and tangent lines? - Solve the two line equations simultaneously.

1. Differentiate: \(f'(x)=2x^3-6x\), \(f''(x)=6x^2-6\), and \(f'''(x)=12x\). 2. Solve \(f''(x)=0\): \(x=-1\) and \(x=1\). Since \(f'''(\pm1)\ne0\), both values give inflection points. 3. The points are \(\left(-1, -\frac{5}{2}\right)\) and \(\left(1, -\frac{5}{2}\right)\). Their tangent slopes are \(f'(-1)=4\) and \(f'(1)=-4\). 4. The tangent lines are \(y=4(x+1)-\frac{5}{2}=4x+\frac{3}{2}\) and \(y=-4(x-1)-\frac{5}{2}=-4x+\frac{3}{2}\). 5. Setting the equations equal gives \(4x+\frac{3}{2}=-4x+\frac{3}{2}\), so \(x=0\). Then \(y=\frac{3}{2}\).

\(S=\left(0, \frac{3}{2}\right)\)

- Use the second and third derivatives to locate and verify the inflection points. - Keep radical values exact. - What must the \(x\)-coordinate be for a point on the \(y\)-axis? - Use the even symmetry of the function when evaluating the two points.

1. Differentiate: \(f'(x)=-x^3+6x\), \(f''(x)=-3x^2+6\), and \(f'''(x)=-6x\). 2. Solve \(f''(x)=0\): \(x=\pm\sqrt{2}\). Since \(f'''(\pm\sqrt{2})\ne0\), both values give inflection points. 3. At both values, \(f(\pm\sqrt{2})=5\). The slopes are \(f'(\sqrt{2})=4\sqrt{2}\) and \(f'(-\sqrt{2})=-4\sqrt{2}\). 4. The tangent lines are \(y=4\sqrt{2}(x-\sqrt{2})+5=4\sqrt{2}x-3\) and \(y=-4\sqrt{2}(x+\sqrt{2})+5=-4\sqrt{2}x-3\). 5. Setting the equations equal gives \(4\sqrt{2}x-3=-4\sqrt{2}x-3\), so \(x=0\). Then \(y=-3\), and therefore \(S=(0, -3)\) lies on the \(y\)-axis.

Tangent lines: \(y=4\sqrt{2}x-3\) and \(y=-4\sqrt{2}x-3\) Intersection: \(S=(0, -3)\), which lies on the \(y\)-axis.

- Concavity is determined by the sign of the second derivative. - The exponential factor is positive for every real input. - Test the sign of the second derivative on the intervals separated by its zeros. - Substitute the inflection-point inputs into the original function.

1. Differentiate: \(\frac{\text{d}}{\text{d}x}\phi(x)=-x\phi(x)\). 2. Differentiate again: \(\frac{\text{d}^2}{\text{d}x^2}\phi(x)=-\phi(x)+x^2\phi(x)=(x^2-1)\phi(x)\). 3. Since \(\phi(x)>0\), the second derivative is zero when \(x^2-1=0\), so \(x=-1\) and \(x=1\). 4. For \(|x|>1\), \(x^2-1>0\), so the graph is concave up. For \(|x|<1\), \(x^2-1<0\), so the graph is concave down. 5. \(\phi(\pm1)=\frac{1}{\sqrt{2\pi}}e^{-1/2}=\frac{1}{\sqrt{2\pi e}}\). Thus, the inflection points are \(\left(-1,\frac{1}{\sqrt{2\pi e}}\right)\) and \(\left(1,\frac{1}{\sqrt{2\pi e}}\right)\).

Concave up on \((-\infty, -1)\) and \((1, \infty)\); concave down on \((-1, 1)\). The inflection points are \(\left(-1, \frac{1}{\sqrt{2\pi e}}\right)\) and \(\left(1, \frac{1}{\sqrt{2\pi e}}\right)\), with \(y\approx0.242\).

- Find the second and third derivatives. - Use the necessary condition for an inflection point. - Verify the result using the third derivative. - Evaluate the derivatives at \(x=0\) without choosing a particular parameter.

1. The first two derivatives are \(h_r'(x)=5rx^4-15x^2\) and \(h_r''(x)=20rx^3-30x\). 2. For an inflection point at \(x=1\), require \(h_r''(1)=20r-30=0\). Thus, \(r=1.5\). The third derivative is \(h_r'''(x)=60rx^2-30\), and \(h_{1.5}'''(1)=60\ne0\), confirming the inflection point. 3. For every real \(r\), \(h_r''(0)=0\) and \(h_r'''(0)=-30\ne0\). Therefore, \(x=0\) is always an inflection point.

a) \(r=1.5\) b) For every \(r\), \(h_r''(0)=0\) and \(h_r'''(0)=-30\ne0\), so \(x=0\) is an inflection point.

- Look for the greatest common exponential factor. - Use the fact that \(e^{2x}=(e^x)^2\). - Set the first derivative equal to zero for extrema and the second derivative equal to zero for possible inflection points. - Verify each type with the next derivative.

1. Factor the function: \(f(x)=2e^x(3-e^x)\). Since \(e^x>0\), the zero satisfies \(e^x=3\), so \(x=\ln(3)\). 2. Differentiate: \(f'(x)=6e^x-4e^{2x}\), \(f''(x)=6e^x-8e^{2x}\), and \(f'''(x)=6e^x-16e^{2x}\). 3. For an extremum, solve \(f'(x)=2e^x(3-2e^x)=0\). Thus \(e^x=\frac{3}{2}\), so \(x=\ln\left(\frac{3}{2}\right)\). The second derivative there is \(-9<0\), so the point is a local maximum. Its y-coordinate is \(\frac{9}{2}\). 4. For an inflection point, solve \(f''(x)=2e^x(3-4e^x)=0\). Thus \(e^x=\frac{3}{4}\), so \(x=\ln\left(\frac{3}{4}\right)\). The third derivative there is \(-\frac{9}{2}\neq0\), confirming an inflection point. Its y-coordinate is \(\frac{27}{8}\).

a) \(f(x)=2e^x(3-e^x)\); zero at \(x=\ln(3)\). b) Local maximum: \(\left(\ln\left(\frac{3}{2}\right),\frac{9}{2}\right)\). Inflection point: \(\left(\ln\left(\frac{3}{4}\right),\frac{27}{8}\right)\).

- Treat the expression as a quadratic in \(e^x\). - Substitute \(u=e^x\), factor, and then substitute back. - Use the first derivative for extrema and the second derivative for possible inflection points. - Verify the classifications and evaluate the original function.

1. Let \(u=e^x\). Then \(g(x)=u^2-4u+3=(u-1)(u-3)\). Substituting back gives \(g(x)=(e^x-1)(e^x-3)\). 2. The zeros satisfy \(e^x=1\) or \(e^x=3\), so \(x=0\) and \(x=\ln(3)\). 3. Differentiate: \(g'(x)=2e^{2x}-4e^x\), \(g''(x)=4e^{2x}-4e^x\), and \(g'''(x)=8e^{2x}-4e^x\). 4. Solve \(g'(x)=2e^x(e^x-2)=0\). Thus \(x=\ln(2)\). Since \(g''(\ln(2))=8>0\), the point is a local minimum, and \(g(\ln(2))=-1\). 5. Solve \(g''(x)=4e^x(e^x-1)=0\). Thus \(x=0\). Since \(g'''(0)=4\neq0\), the point is an inflection point, and \(g(0)=0\).

a) \(g(x)=(e^x-1)(e^x-3)\); zeros at \(x=0\) and \(x=\ln(3)\). b) Local minimum: \((\ln(2),-1)\). Inflection point: \((0,0)\).

- Substitute \(u=e^x\) to turn the zero equation into a quadratic. - Use the first derivative for the local extremum. - Use the second derivative for a possible inflection point. - Remember that \(e^x\) is never zero.

1. Let \(u=e^x\). The zero equation becomes \(u^2-5u+4=0\), or \((u-1)(u-4)=0\). Thus \(e^x=1\) or \(e^x=4\), giving \(x=0\) and \(x=\ln(4)\). 2. Differentiate: \(f'(x)=2e^{2x}-5e^x=e^x(2e^x-5)\). The critical number satisfies \(e^x=\frac{5}{2}\), so \(x=\ln\left(\frac{5}{2}\right)\). 3. The second derivative is \(f''(x)=4e^{2x}-5e^x=e^x(4e^x-5)\). At the critical number, \(f''(x)=\frac{25}{2}>0\), so the point is a local minimum. Its y-coordinate is \(-\frac{9}{4}\). 4. A possible inflection point satisfies \(f''(x)=0\), so \(e^x=\frac{5}{4}\) and \(x=\ln\left(\frac{5}{4}\right)\). The third derivative is \(f'''(x)=8e^{2x}-5e^x\), which equals \(\frac{25}{4}\neq0\) there. Therefore, the graph has an inflection point at that x-coordinate.

a) \(x=0\) and \(x=\ln(4)\). b) Local minimum: \(\left(\ln\left(\frac{5}{2}\right),-\frac{9}{4}\right)\). c) Inflection-point x-coordinate: \(x=\ln\left(\frac{5}{4}\right)\).

- Exponential expressions are always positive. - Compare \(f(-x)\) with \(f(x)\). - Apply the chain rule to both exponential terms. - The sign of the second derivative determines concavity.

1. Since \(f(0)=1+1=2\), the y-intercept is \((0,2)\). Both exponential terms are positive for every real \(x\), so their sum cannot be zero. Therefore, there are no x-intercepts. 2. \(f(-x)=e^{-0.2x}+e^{0.2x}=f(x)\), so \(f\) is even. As \(x\to\infty\), the first term grows without bound and the second approaches \(0\), so \(f(x)\to\infty\). By symmetry, \(f(x)\to\infty\) as \(x\to-\infty\). 3. The derivatives are \(f'(x)=0.2e^{0.2x}-0.2e^{-0.2x}\) and \(f''(x)=0.04e^{0.2x}+0.04e^{-0.2x}=0.04f(x)\). Since \(f(x)>0\), \(f''(x)>0\) for all \(x\), so the graph is concave up everywhere. 4. Set \(f'(x)=0\): \(e^{0.2x}=e^{-0.2x}\), so \(x=0\). Because the graph is concave up, \((0,2)\) is a local minimum.

a) Y-intercept: \((0,2)\). There are no x-intercepts because \(e^{0.2x}>0\) and \(e^{-0.2x}>0\). b) \(f\) is even. Also, \(\lim_{x\to\infty}f(x)=\infty\) and \(\lim_{x\to-\infty}f(x)=\infty\). c) \(f''(x)=0.04f(x)>0\), so the graph is concave up for all real \(x\). d) Local minimum: \((0,2)\).

- Compare \(f(-x)\) with \(f(x)\). - Use the product rule and chain rule when differentiating. - The exponential factor is always positive. - Use the sign of the second derivative to determine concavity.

1. Since \(f(-x)=4e^{-(-x)^2/2}=f(x)\), the function is even. Also, \(-x^2/2\to-\infty\) as \(x\to\pm\infty\), so \(f(x)\to0\). 2. The derivatives are \(f'(x)=-4xe^{-x^2/2}\) and \(f''(x)=(4x^2-4)e^{-x^2/2}\). The first derivative is zero only at \(x=0\), and \(f''(0)=-4<0\). Thus \((0,4)\) is a local maximum. 3. Since the exponential factor is positive, \(f''(x)=0\) when \(x^2-1=0\), so the inflection-point x-coordinates are \(x=-1\) and \(x=1\). The second derivative is positive for \(|x|>1\) and negative for \(|x|<1\). Therefore, the graph is concave up on \((-\infty,-1)\cup(1,\infty)\) and concave down on \((-1,1)\).

a) \(f\) is even, and \(\lim_{x\to\pm\infty}f(x)=0\). b) Local maximum: \((0,4)\). c) Inflection-point x-coordinates: \(x=-1\) and \(x=1\). The graph is concave up on \((-\infty,-1)\cup(1,\infty)\) and concave down on \((-1,1)\).

- Apply both the product and chain rules. - The exponential factor is always positive. - Solve each resulting quadratic equation. - Use \(k>0\) to verify that the square roots are real.

1. Apply the product and chain rules: \(f_k'(x)=(-x^2+2x+k)e^{-x}\) and \(f_k''(x)=(x^2-4x+2-k)e^{-x}\). 2. Since \(e^{-x}>0\), the critical numbers satisfy \(x^2-2x-k=0\). 3. Solving gives \(x=1\pm\sqrt{1+k}\). 4. The x-coordinates of inflection points satisfy \(x^2-4x+2-k=0\). 5. Solving gives \(x=2\pm\sqrt{2+k}\). Because \(k>0\), the two roots are real and distinct. The quadratic factor in \(f_k''\) changes sign at each root, so both are x-coordinates of inflection points.

Critical numbers: \(x=1\pm\sqrt{1+k}\) Inflection-point x-coordinates: \(x=2\pm\sqrt{2+k}\)

- The nonzero parameter does not affect where the derivative is zero. - Factor the cubic expression before solving it. - Check the sign changes required for local extrema and inflection points. - Decide whether the requested x-values depend on \(a\).

1. Differentiate: \(g_a'(x)=a(1-x^2)e^{-x^2/2}\). 2. Since \(a\neq0\) and the exponential factor is positive, the critical numbers satisfy \(1-x^2=0\). Thus, \(x=\pm1\). 3. Differentiate again: \(g_a''(x)=a(x^3-3x)e^{-x^2/2}\). 4. The x-coordinates of inflection points satisfy \(x(x^2-3)=0\), so \(x=0\) and \(x=\pm\sqrt{3}\). 5. The relevant derivative factors change sign at each listed value, confirming the local extrema and inflection points.

Critical numbers: \(x=-1\) and \(x=1\) Inflection-point x-coordinates: \(x=-\sqrt{3}\), \(x=0\), and \(x=\sqrt{3}\)

- Compare \(f(-x)\) with \(f(x)\) and \(-f(x)\). - Analyze the quadratic exponent and use the fact that \(e^u\) increases as \(u\) increases. - Inflection points occur where the second derivative changes sign. - Use point-slope form for the tangent line.

1. Since \(f(-x)=2e^{1-(-x)^2/2}=f(x)\), the function is even, so its graph is symmetric about the y-axis. It is not symmetric about the origin. As \(x\to\pm\infty\), the exponent approaches \(-\infty\), so \(f(x)\to0\). 2. The exponent \(1-x^2/2\) has its maximum value \(1\) at \(x=0\). Because the exponential function is strictly increasing, \(f\) is also largest at \(x=0\). Thus the maximum point is \((0,2e)\). 3. Differentiate: \(f'(x)=-2xe^{1-x^2/2}\), and \(f''(x)=(2x^2-2)e^{1-x^2/2}\). Since the exponential factor is positive, the second derivative is zero at \(x=\pm1\). It changes sign at both values, so the inflection points are \((-1,2\sqrt{e})\) and \((1,2\sqrt{e})\). 4. At \(x=1\), \(f(1)=2\sqrt{e}\) and \(f'(1)=-2\sqrt{e}\). Using point-slope form, \(y-2\sqrt{e}=-2\sqrt{e}(x-1)\), or \(y=-2\sqrt{e}x+4\sqrt{e}\).

a) The graph is symmetric about the y-axis, and \(\lim_{x\to\pm\infty}f(x)=0\). b) Maximum: \((0,2e)\). c) Inflection points: \((-1,2\sqrt{e})\) and \((1,2\sqrt{e})\). d) \(y=-2\sqrt{e}x+4\sqrt{e}\).

- Differentiate and factor the exponential terms. - How do zeros of a derivative identify extrema? - What happens to \(e^x\) as \(x\to-\infty\)? - How are zeros of a function related to extrema of its antiderivative? - How are extrema of a derivative related to inflection points of the original function?

1. Differentiate and factor: \(f'(x)=4e^x-2e^{2x}=2e^x(2-e^x)\). 2. Set \(f'(x)=0\). Since \(e^x>0\), \(e^x=2\), so \(x=\ln(2)\). Also, \(f''(\ln(2))=4(2)-4(4)=-8<0\), so \(f\) has a local maximum there. Its value is \(f(\ln(2))=8-4=4\). 3. As \(x\to-\infty\), both \(e^x\) and \(e^{2x}\) approach \(0\). Therefore, \(F(x)=4e^x-\frac{1}{2}e^{2x}\to0\). 4. Because \(F'=f\), the sign change of \(f\) from positive to negative at \(x=\ln(4)\) gives a local maximum of \(F\). 5. Because \(F''=f'\), the local maximum of \(f\) at \(x=\ln(2)\) corresponds to an inflection point of \(F\). Its y-coordinate is \(F(\ln(2))=4(2)-\frac{1}{2}(4)=6\).

a) \(f'(x)=4e^x-2e^{2x}=2e^x(2-e^x)\) b) Local maximum at \((\ln(2),4)\) c) \(\lim_{x\to-\infty}F(x)=0\) d) \(F\) has a local maximum at \(x=\ln(4)\) and an inflection point at \((\ln(2),6)\).

- Compare linear growth with exponential decay. - Set the first derivative equal to zero and check whether \(c\) remains in the resulting equation. - The greatest decrease occurs where the first derivative is minimal. - Substitute the maximum time into the original model.

1. Since exponential decay dominates linear growth, \(h_c(t)=\frac{c(t+40)}{e^{0.02t}}\to0\) as \(t\to\infty\). 2. Differentiate: \(h_c'(t)=ce^{-0.02t}(0.2-0.02t)\). The derivative is zero at \(t=10\), independent of \(c\), and changes from positive to negative there. 3. The second derivative is \(h_c''(t)=ce^{-0.02t}(0.0004t-0.024)\). It is zero at \(t=60\). At that time the first derivative reaches its minimum, so the concentration is decreasing most rapidly. 4. Use the maximum at \(t=10\): \(c(50)e^{-0.2}=500\). Thus, \(c=10e^{0.2}\approx12.21\,\text{mg}/(\text{L}\cdot\text{h})\).

a) \(0\,\text{mg/L}\) b) \(t=10\) hours c) \(t=60\) hours d) \(c=10e^{0.2}\approx12.21\,\text{mg}/(\text{L}\cdot\text{h})\)

- What equations follow from the origin being an inflection point on the graph? - At the point of tangency, the function and tangent line have the same value and slope. - Use the inflection-point conditions to reduce the number of unknown coefficients. - Solve the resulting linear system.

1. Write a general cubic polynomial: \(f(x)=ax^3+bx^2+cx+d\). 2. Since the origin is on the graph, \(f(0)=0\), so \(d=0\). Also, \(f''(x)=6ax+2b\), and the inflection point at \(x=0\) gives \(f''(0)=0\), so \(b=0\). Thus \(f(x)=ax^3+cx\). 3. The tangent line has value \(-3(2)+10=4\) at \(x=2\), so \(f(2)=4\). Its slope is \(-3\), so \(f'(2)=-3\). 4. These conditions give \(8a+2c=4\) and \(12a+c=-3\). 5. Solving the system gives \(a=-\frac{5}{8}\) and \(c=\frac{9}{2}\). 6. Therefore, \(f(x)=-\frac{5}{8}x^3+\frac{9}{2}x\). Since \(f'''(x)=-\frac{15}{4}\neq0\), the origin is indeed an inflection point.

\(f(x)=-\frac{5}{8}x^3+\frac{9}{2}x\)

- Use the center point to find \(A\). - Substitute the second known point to solve for \(k\). - The most extreme slope occurs at an inflection point. - Multiply the decimal slope magnitude by \(100\%\) to obtain percent grade.

1. Since \(f(0)=A=12\), \(A=12\). Using \(f(20)=8\), \(12e^{-400k}=8\), so \(k=\frac{\ln(1.5)}{400}\approx0.001014\,\text{m}^{-2}\). 2. The steepest slope occurs at an inflection point. Since \(f''(x)=12e^{-kx^2}(-2k+4k^2x^2)\), the positive inflection point satisfies \(x^2=\frac{1}{2k}\). Thus, \(x=\sqrt{\frac{200}{\ln(1.5)}}\approx22.21\). At this point, \(f(x)=12e^{-1/2}\approx7.28\). The point is approximately \((22.21, 7.28)\). 3. The first derivative is \(f'(x)=-2kxf(x)\). At the inflection point, \(f'(x)\approx-0.328\), which corresponds to a downhill grade of about \(32.8\%\).

a) \(A=12\), \(k=\frac{\ln(1.5)}{400}\approx0.001014\,\text{m}^{-2}\) b) Approximately \((22.21, 7.28)\) c) About \(32.8\%\) downhill

- Use y-axis symmetry to determine which powers of \(x\) may appear. - Translate the intercept information into equations. - Use the second derivative for the inflection-point condition. - Solve the resulting system for the coefficients.

1. Symmetry about the y-axis means the polynomial has only even powers, so write \(f(x)=ax^4+bx^2+c\). 2. The y-intercept gives \(f(0)=4\), so \(c=4\). 3. The x-intercept gives \(f(2)=16a+4b+4=0\). 4. Since \(f''(x)=12ax^2+2b\), the inflection-point condition \(f''(1)=0\) gives \(12a+2b=0\), or \(b=-6a\). 5. Substitute into the intercept equation: \(16a+4(-6a)+4=0\), so \(a=\frac12\) and \(b=-3\). 6. Thus, \(f(x)=\frac12x^4-3x^2+4\). Also, \(f'''(1)=12\ne0\), confirming an inflection point at \(x=1\).

\(f(x)=\frac12x^4-3x^2+4\)

- Identify which factor can equal zero. - Compare linear growth with exponential growth or decay. - Use the product and chain rules. - An inflection point occurs where the second derivative changes sign. - Use the point and slope to write the tangent line.

1. Since the exponential factor is never zero, \(f(x)=0\) only when \(x-2=0\). Thus the zero is \(x=2\). 2. As \(x\to\infty\), exponential decay dominates the linear factor, so \(f(x)\to0\). As \(x\to-\infty\), \(x-2\to-\infty\) and \(e^{-0.5x}\to\infty\), so \(f(x)\to-\infty\). 3. The first derivative is \(f'(x)=(2-0.5x)e^{-0.5x}\). It is zero at \(x=4\). The second derivative is \(f''(x)=(0.25x-1.5)e^{-0.5x}\), and \(f''(4)=-0.5e^{-2}<0\). Therefore, the local maximum is \((4,2e^{-2})\). 4. The second derivative is zero only at \(x=6\), and its linear factor changes sign there. Thus the unique inflection point is \((6,4e^{-3})\). The slope there is \(f'(6)=-e^{-3}\). Using point-slope form gives \(y-4e^{-3}=-e^{-3}(x-6)\), or \(y=-e^{-3}x+10e^{-3}\).

a) \(x=2\). b) \(\lim_{x\to\infty}f(x)=0\), and \(\lim_{x\to-\infty}f(x)=-\infty\). c) Local maximum: \((4,2e^{-2})\). d) Inflection point: \((6,4e^{-3})\). Tangent line: \(y=-e^{-3}x+10e^{-3}\).

- An exponential factor is never zero. - Use the product and chain rules. - A simple zero of the second derivative produces a sign change here. - Compare linear growth with exponential growth or decay.

1. Since the exponential factor is never zero, \(f(x)=0\) only when \(2x=0\). Thus the x-intercept is \((0,0)\). 2. Differentiate: \(f'(x)=(2-2x)e^{1-x}\). The derivative is zero at \(x=1\). The second derivative is \(f''(x)=(2x-4)e^{1-x}\), and \(f''(1)=-2<0\). Therefore, the local maximum is \((1,2)\). 3. The second derivative is zero only at \(x=2\). Its linear factor changes sign there, so this is the unique inflection point. Since \(f(2)=4e^{-1}\), the point is \(\left(2,\frac{4}{e}\right)\). 4. As \(x\to\infty\), exponential decay dominates linear growth, so \(f(x)\to0\). As \(x\to-\infty\), the linear factor is negative and unbounded while \(e^{1-x}\to\infty\), so \(f(x)\to-\infty\).

a) \((0,0)\). b) Local maximum: \((1,2)\). c) Inflection point: \(\left(2,\frac{4}{e}\right)\). d) \(\lim_{x\to\infty}f(x)=0\), and \(\lim_{x\to-\infty}f(x)=-\infty\).

- Use the leading term to determine end behavior. - Inflection points occur where the second derivative changes sign. - A point on the y-axis has x-coordinate zero. - Evaluate the first derivative at the inflection point to find the tangent slope.

1. The leading term is \(\frac13kx^3\). Therefore, when \(k>0\), \(f_k(x)\to\infty\) as \(x\to\infty\) and \(f_k(x)\to-\infty\) as \(x\to-\infty\). When \(k<0\), these directions reverse. 2. Differentiate twice: \(f_k'(x)=kx^2+2(k+2)x+12\), \(f_k''(x)=2kx+2k+4\). Setting the second derivative equal to zero gives \(2kx+2k+4=0\), so \(x=-1-\frac2k\). Since \(f_k'''(x)=2k\ne0\), this critical point of the first derivative is an inflection point. 3. For the inflection point to lie on the y-axis, \(-1-\frac2k=0\), which gives \(k=-2\). Then \(f_{-2}(x)=-\frac23x^3+12x\), so \(f_{-2}(0)=0\), and the inflection point is \((0, 0)\). Finally, \(f_{-2}'(x)=-2x^2+12\), so the tangent slope at the origin is \(f_{-2}'(0)=12\).

a) For \(k>0\), \(f_k(x)\to\infty\) as \(x\to\infty\) and \(f_k(x)\to-\infty\) as \(x\to-\infty\). For \(k<0\), \(f_k(x)\to-\infty\) as \(x\to\infty\) and \(f_k(x)\to\infty\) as \(x\to-\infty\). b) \(x=-1-\frac2k\) c) \(k=-2\); \(W_{-2}=(0, 0)\); tangent slope \(12\).

- Use y-axis symmetry to simplify the general fourth-degree polynomial. - Translate the y-intercept, point, slope, and inflection-point information into equations. - Solve the resulting system for the coefficients. - Verify the inflection point with the third derivative.

1. Symmetry about the y-axis gives \(f(x)=ax^4+bx^2+c\). The y-intercept gives \(c=2.5=\frac52\). 2. The point, slope, and inflection-point conditions at \(x=1\) give \(a+b+\frac52=0\), \(4a+2b=-4\), and \(12a+2b=0\). 3. The third equation gives \(b=-6a\). Substitute into the second equation: \(4a-12a=-4\), so \(a=\frac12\) and \(b=-3\). Therefore, \(f(x)=\frac12x^4-3x^2+\frac52\). The point condition is satisfied, and \(f'''(1)=12\ne0\), confirming the inflection point.

\(f(x)=\frac12x^4-3x^2+\frac52\)

- Look for an input that makes the function value independent of \(k\). - Use the product and chain rules to find the first two derivatives. - Eliminate \(k\) from the coordinates of the local maxima.

1. At \(x=0\), \(f_k(0)=0\) for every \(k\), so \((0, 0)\) is common to all graphs. If two distinct parameter values \(k\) and \(m\) give the same function value at \(x\ne0\), then \(e^{1-kx}=e^{1-mx}\), which implies \((k-m)x=0\), a contradiction. Thus, there is no other common point. 2. Differentiate: \(f_k'(x)=(1-kx)e^{1-kx}\) and \(f_k''(x)=(k^2x-2k)e^{1-kx}\). The first derivative is zero at \(x=\frac1k\). Since \(f_k''\left(\frac1k\right)=-k<0\), the point is a local maximum, with \(H=\left(\frac1k, \frac1k\right)\). The second derivative is zero at \(x=\frac2k\). Its sign changes there, so the inflection point is \(W=\left(\frac2k, \frac{2}{ke}\right)\). 3. Every local maximum has equal x- and y-coordinates, so all local maxima lie on \(y=x\).

a) \((0, 0)\) b) \(H=\left(\frac1k, \frac1k\right)\) and \(W=\left(\frac2k, \frac{2}{ke}\right)\) c) \(y=x\)

- Find where the denominator is zero and use the numerator discriminant to check for x-intercepts. - Rewrite the numerator using \((x-2)^2\). - Use the sign of the second derivative to determine concavity. - An inflection point must be a point on the graph where concavity changes.

1. The denominator is zero at \(x=2\), so the domain is \(\mathbb{R}\setminus\{2\}\). 2. The numerator has discriminant \((-4)^2-4\cdot 1\cdot 7=-12<0\), so it has no real zeros. Therefore, the graph has no x-intercepts. 3. Since \((x-2)^2+3=x^2-4x+7\), \(f(x)=x-2+\frac{3}{x-2}\). 4. The vertical asymptote is \(x=2\), and the slant asymptote is \(y=x-2\). 5. For \(x<2\), \((x-2)^3<0\), so \(f''(x)<0\), and the graph is concave down. For \(x>2\), \(f''(x)>0\), and the graph is concave up. 6. Although concavity differs across \(x=2\), that value is not in the domain. Also, \(f''(x)\) is never zero. Therefore, the graph has no inflection points.

a) Domain: \(\mathbb{R}\setminus\{2\}\); no x-intercepts b) Vertical asymptote: \(x=2\); slant asymptote: \(y=x-2\) c) Concave down on \((-\infty, 2)\), concave up on \((2, \infty)\), and no inflection points

- Apply the chain rule to the logarithm. - Analyze the signs of the numerator and denominator. - Use the quotient rule for the second derivative. - No change in concavity means no inflection point.

1. Apply the chain rule: \(g'(x)=\frac{e^x}{e^x+5}\). Both the numerator and denominator are positive, so \(g'(x)>0\) for every real \(x\). Therefore, \(g\) is strictly increasing and has no local extrema. 2. Differentiate using the quotient rule: \(g''(x)=\frac{e^x(e^x+5)-e^x\cdot e^x}{(e^x+5)^2}=\frac{5e^x}{(e^x+5)^2}\). 3. The second derivative is positive for every real \(x\), so the graph is concave up everywhere. Since concavity never changes, the graph has no inflection points.

\(g'(x)=\frac{e^x}{e^x+5}>0\), so \(g\) is strictly increasing and has no local extrema. Also, \(g''(x)=\frac{5e^x}{(e^x+5)^2}>0\), so the graph is always concave up and has no inflection points.

- Inflection points are investigated with the second derivative. - Treat \(k\) as a nonzero constant. - Verify a sign change in the second derivative.

1. The first derivative is \(h_k'(x)=\frac{2kx}{x^2+1}\). 2. Differentiating again gives \(h_k''(x)=\frac{2k(1-x^2)}{(x^2+1)^2}\). 3. Since \(k\ne0\) and the denominator is positive, \(h_k''(x)=0\) exactly when \(1-x^2=0\), so \(x=\pm1\). 4. The factor \(1-x^2\) changes sign at both values, so both are inflection-point x-coordinates. Neither depends on \(k\).

\(x=-1\) and \(x=1\)

- Check whether the denominator can equal zero. - Compare \(f(-x)\) with \(f(x)\), and compare the leading terms for end behavior. - Use the sign of the first derivative to classify the critical point. - Set the second derivative equal to zero and verify a sign change.

1. Since \(x^2+2>0\) for every real \(x\), the domain is \(\mathbb{R}\). Setting the numerator equal to zero gives \(8-2x^2=0\), so the x-intercepts are \((-2, 0)\) and \((2, 0)\). 2. Since \(f(-x)=f(x)\), the graph is symmetric about the y-axis. The numerator and denominator have the same degree, so \(\lim_{x\to\pm\infty}f(x)=-2\). 3. The first derivative is \(f'(x)=\frac{-24x}{(x^2+2)^2}\). It is positive for \(x<0\) and negative for \(x>0\), so the graph has a local maximum at \((0, 4)\). 4. The second derivative is \(f''(x)=\frac{72x^2-48}{(x^2+2)^3}=\frac{24(3x^2-2)}{(x^2+2)^3}\). It is zero when \(x=\pm\sqrt{\frac{2}{3}}=\pm\frac{\sqrt{6}}{3}\), and its sign changes at both values. 5. Substitution gives \(f\left(\pm\frac{\sqrt{6}}{3}\right)=\frac{5}{2}\). Therefore, the inflection points are \(\left(-\frac{\sqrt{6}}{3}, \frac{5}{2}\right)\) and \(\left(\frac{\sqrt{6}}{3}, \frac{5}{2}\right)\).

a) Domain: \(\mathbb{R}\); x-intercepts: \((-2, 0)\) and \((2, 0)\) b) Symmetric about the y-axis; \(\lim_{x\to\pm\infty}f(x)=-2\) c) Local maximum at \((0, 4)\) d) Inflection points: \(\left(-\frac{\sqrt{6}}{3}, \frac{5}{2}\right)\) and \(\left(\frac{\sqrt{6}}{3}, \frac{5}{2}\right)\)

- Which values make the denominator zero? - Use the quotient rule or rewrite the function with a negative exponent. - Relate the sign of the second derivative to concavity. - A sign change can occur at a point where the second derivative is undefined. - What must exist for an inflection point to exist?

1. The denominator is zero at \(x=-1\) and \(x=1\), so the domain is \(\mathbb{R}\setminus\{-1, 1\}\). Differentiating gives \(f'(x)=-2x(x^2-1)^{-2}\) and \(f''(x)=\frac{6x^2+2}{(x^2-1)^3}\). 2. The numerator satisfies \(6x^2+2\ge2>0\), so \(f''(x)\) cannot equal zero. 3. When \(|x|>1\), the denominator \((x^2-1)^3\) is positive, so \(f''(x)>0\) and the graph is concave up. When \(|x|<1\), the denominator is negative, so \(f''(x)<0\) and the graph is concave down. 4. The factor \(x^2-1\) changes sign across each excluded value, causing adjacent graph branches to have different concavity. However, the function is not defined at \(x=-1\) or \(x=1\), so no point of the graph—and therefore no inflection point—exists there.

1. Domain: \(\mathbb{R}\setminus\{-1, 1\}\); \(f''(x)=\frac{6x^2+2}{(x^2-1)^3}\) 2. \(f''\) has no zeros because \(6x^2+2>0\) for every real \(x\). 3. Concave up on \((-\infty, -1)\) and \((1, \infty)\); concave down on \((-1, 1)\) 4. The concavity differs across the vertical asymptotes, but the excluded values are not points on the graph.

- Evaluate the model at \(t=0\) and examine the exponential term as \(t\) becomes large. - Half the population corresponds to \(N(t)=1000\). - Differentiate using the chain rule. - A maximum growth rate corresponds to a change in concavity of the original function.

1. \(N(0)=\frac{2000}{40}=50\), so \(50\) people initially know the message. Since \(e^{-0.4t}\to 0\), \(\lim_{t\to\infty}N(t)=2000\), meaning the model approaches the city's full population. 2. Set \(N(t_0)=1000\). Then \(1+39e^{-0.4t_0}=2\), so \(e^{0.4t_0}=39\). Therefore, \(t_0=\frac{\ln 39}{0.4}\approx 9.16\) hours. 3. Differentiate: \(N'(t)=\frac{31{,}200e^{-0.4t}}{(1+39e^{-0.4t})^2}\). Thus, \(N'(0)=\frac{31{,}200}{1600}=19.5\) people per hour. 4. Because \(N'(t)\) is greatest at \(t_0\), the graph of \(N\) has an inflection point there. Equivalently, the concavity changes and \(N''(t_0)=0\).

a) \(N(0)=50\); \(\lim_{t\to\infty}N(t)=2000\) b) \(t_0=\frac{\ln 39}{0.4}\approx 9.16\,\text{hours}\) c) \(N'(0)=19.5\) people per hour d) An inflection point, because the first derivative has a maximum there

- Use the signs of the first and second derivatives. - Analyze the intervals separated by the excluded domain value. - A vertical asymptote occurs when the denominator approaches zero while the numerator remains nonzero. - Check the denominators of the derivative formulas.

1. The derivatives are \(f'(x)=\frac{-3}{(x+1)^2}\) and \(f''(x)=\frac{6}{(x+1)^3}\). 2. Since \((x+1)^2>0\) on the domain, \(f'(x)<0\). Therefore, \(f\) is strictly decreasing on \((-\infty,-1)\) and on \((-1,\infty)\). 3. For \(x<-1\), \((x+1)^3<0\), so \(f''(x)<0\) and the graph is concave down. For \(x>-1\), \(f''(x)>0\) and the graph is concave up. 4. The denominator of \(f\) is zero at \(x=-1\), while the numerator is \(3\neq0\), so \(x=-1\) is a vertical asymptote. The derivatives have nonzero constant numerators and denominators containing powers of \(x+1\), so their magnitudes also become unbounded as \(x\to-1\). Thus \(x=-1\) is the common vertical asymptote.

1. Strictly decreasing on \((-\infty,-1)\) and \((-1,\infty)\). Concave down on \((-\infty,-1)\); concave up on \((-1,\infty)\). 2. The graphs of \(f\), \(f'\), and \(f''\) all have vertical asymptote \(x=-1\).

- Determine the degree of the first derivative. - Local extrema require critical values. - Determine the degree of the second derivative. - After finding an upper bound, show that a sixth-degree polynomial can actually attain it.

1. The first derivative of a sixth-degree polynomial has degree \(5\), so it has at most \(5\) real zeros. Since every local extremum must occur at a critical value, \(f\) can have at most \(5\) local extrema. 2. The second derivative has degree \(4\), so it has at most \(4\) real zeros. Because \(f''\) is continuous, every inflection point must occur at a zero of \(f''\). Thus \(f\) can have at most \(4\) inflection points. 3. This bound is attainable. For example, choose \(f''(x)=(x^2-1)(x^2-4)\). Its four simple zeros are \(x=-2,-1,1,2\), and \(f''\) changes sign at each one. Integrating twice gives a sixth-degree polynomial, for example \(f(x)=\frac{1}{30}x^6-\frac{5}{12}x^4+2x^2\). Therefore, the greatest possible number of inflection points is \(4\).

a) At most \(5\) local extrema b) \(4\) inflection points

- Factor the first derivative to analyze its sign. - No sign change in the first derivative means no local extremum. - To classify the non-extremum horizontal tangent, compute the second derivative and check whether concavity changes.

1. Differentiate: \(g'(x)=4x^3-6x^2=2x^2(2x-3)\), so \(g'(0)=0\). 2. Near \(x=0\), the factor \(2x^2\) is positive for \(x\neq0\), while \(2x-3\) is negative. Thus \(g'(x)<0\) on both sides of \(0\). 3. Because the derivative does not change sign, there is no local extremum at \(x=0\). 4. Differentiate again: \(g''(x)=12x^2-12x=12x(x-1)\). For \(x<0\) close to \(0\), \(g''(x)>0\); for \(0<x<1\), \(g''(x)<0\). The concavity changes from up to down at \(x=0\). Together with \(g'(0)=0\), this shows that \(x=0\) is a stationary inflection point.

There is no local extremum at \(x=0\) because \(g'(x)\) is negative on both sides. The point is a stationary inflection point.

- Find the second derivative of each function. - A continuous function cannot change sign at a point where its value is nonzero. - Examine the sign immediately to the left and right of each specified input. - A linear expression changes sign at its zero when its slope is nonzero.

1. For part a, \(g''(x)=-\sin x\). Near \(x=0\), \(-\sin x\) is positive to the left and negative to the right, so the second derivative changes sign at \(0\). At \(x=\frac{\pi}{2}\), \(g''\left(\frac{\pi}{2}\right)=-1\ne0\). Because the second derivative is continuous and nonzero there, it does not change sign at \(\frac{\pi}{2}\). 2. For part b, \(h'(x)=3x^2-12x+12\) and \(h''(x)=6x-12=6(x-2)\). This expression is negative for \(x<2\) and positive for \(x>2\), so the second derivative changes sign at \(x=2\).

a) At \(x=0\): yes. At \(x=\frac{\pi}{2}\): no. b) At \(x=2\): yes.

- A sufficient condition belongs in the hypothesis of an if-then statement. - Check whether the second derivative changes sign for \(x^4\). - Use continuity and the intermediate value theorem.

1. If \(f''(x_0)=0\) and \(f'''(x_0)\neq0\), then the graph of \(f\) has an inflection point at \(x_0\). 2. For \(f(x)=x^4\), \(f''(x)=12x^2\), so \(f''(0)=0\). However, \(f''(x)>0\) on both sides of \(0\), so there is no change in concavity. The condition is not sufficient. 3. At an inflection point, the continuous second derivative changes sign. By continuity, it must pass through \(0\), so \(f''(x_0)=0\) is necessary.

a) If \(f''(x_0)=0\) and \(f'''(x_0)\neq0\), then \(f\) has an inflection point at \(x_0\). b) No. The function \(f(x)=x^4\) has \(f''(0)=0\) but no inflection point. c) A continuous second derivative that changes sign must equal \(0\) at the change point.

- Relate the second derivative to concavity. - First find the values where the second derivative can change sign. - Test the sign of the second derivative on each interval. - What do positive and negative second-derivative values mean?

1. Differentiate: \(f'(x)=\frac{1}{4}x^4-\frac{1}{2}x^2\) and \(f''(x)=x^3-x=x(x-1)(x+1)\). 2. The zeros of \(f''\) are \(-1\), \(0\), and \(1\). 3. A sign chart gives \(f''(x)<0\) on \((-\infty, -1)\), \(f''(x)>0\) on \((-1, 0)\), \(f''(x)<0\) on \((0, 1)\), and \(f''(x)>0\) on \((1, \infty)\). 4. Therefore, the graph is concave up where \(f''>0\) and concave down where \(f''<0\).

Concave up on \((-1, 0)\cup(1, \infty)\) Concave down on \((-\infty, -1)\cup(0, 1)\)

- Use the second derivative to analyze concavity. - Solve for the zeros of the second derivative. - Divide the domain at those inputs. - Test the sign of the second derivative on each interval.

1. Differentiate twice: \(f'(x)=2x+4\cos x\) and \(f''(x)=2-4\sin x\). 2. Find where the second derivative is zero: \(2-4\sin x=0\), so \(\sin x=\frac{1}{2}\). On \([0, 2\pi]\), this occurs at \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\). 3. Test the resulting intervals. The second derivative is positive on \(\left(0, \frac{\pi}{6}\right)\) and \(\left(\frac{5\pi}{6}, 2\pi\right)\), so the graph is concave up there. It is negative on \(\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\), so the graph is concave down there.

Concave up on \(\left(0, \frac{\pi}{6}\right)\cup\left(\frac{5\pi}{6}, 2\pi\right)\) Concave down on \(\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\)

- What must the first and second derivatives equal at a stationary inflection point? - First find the inflection value, which does not depend on \(k\). - Require the tangent slope to be zero at that value. - Substitute the resulting parameter into the function to find the point.

1. Differentiate: \(g_k'(x)=x^2+2x+k\), \(g_k''(x)=2x+2\), and \(g_k'''(x)=2\). 2. Solve \(g_k''(x)=0\): \(2x+2=0\), so the inflection value is \(x=-1\). Since \(g_k'''(-1)=2\ne0\), this is an inflection value for every \(k\). 3. For the tangent to be horizontal, require \(g_k'(-1)=0\). Then \(1-2+k=0\), so \(k=1\). 4. Evaluate \(g_1(-1)=-\frac{1}{3}+1-1=-\frac{1}{3}\).

\(k=1\), with stationary inflection point \(\left(-1, -\frac{1}{3}\right)\)

- Which derivative is used to locate possible inflection points? - How many real zeros can the resulting quadratic have? - Why is finding zeros alone not enough? - How does the graph of the second derivative show a sign change at a simple zero?

1. Differentiate twice: \(f'(x)=\frac{1}{24}(4x^3-6x^2-72x)\) and \(f''(x)=\frac{1}{24}(12x^2-12x-72)=\frac{1}{2}(x^2-x-6)\). 2. Factor the second derivative: \(f''(x)=\frac{1}{2}(x-3)(x+2)\). Its only zeros are \(x=-2\) and \(x=3\). 3. Both zeros are simple, so \(f''\) changes sign at each one. Equivalently, \(f'''(x)=x-\frac{1}{2}\), and \(f'''(-2)=-\frac{5}{2}\ne0\) while \(f'''(3)=\frac{5}{2}\ne0\). 4. Therefore, the graph has inflection points at exactly those two \(x\)-values and no others.

Exactly two inflection points, with \(x\)-coordinates \(-2\) and \(3\)

- What must happen to the second derivative when concavity changes? - How can a discriminant determine the number of real zeros of a quadratic? - Why does a simple zero of the second derivative produce a sign change?

1. Differentiate twice: \(g'(x)=0.4x^3+2.4x^2-1.2x\) and \(g''(x)=1.2x^2+4.8x-1.2=1.2(x^2+4x-1)\). 2. Solve \(x^2+4x-1=0\). The discriminant is \(4^2-4(1)(-1)=20>0\), so there are two distinct real zeros, \(x=-2\pm\sqrt{5}\). 3. Both zeros are simple zeros of the quadratic \(g''\), so the second derivative changes sign at each one. 4. Since a quadratic has no other zeros, the graph changes concavity at exactly those two values.

The graph changes concavity at exactly \(x=-2-\sqrt{5}\) and \(x=-2+\sqrt{5}\).

- Relate the sign of the second derivative to concavity. - Where can the second derivative change sign? - What happens to \(f''\) immediately on either side of each zero? - What is the second derivative of a line?

1. Differentiate twice: \(f'(x)=\frac{1}{3}x^3-x\) and \(f''(x)=x^2-1\). 2. The zeros of \(f''\) are \(x=-1\) and \(x=1\). 3. The second derivative is positive for \(x<-1\) and \(x>1\), so the graph is concave up there. It is negative for \(-1<x<1\), so the graph is concave down there. 4. At each transition value, \(f''=0\) only at the single point and changes sign across it. For a line, the second derivative is zero at every point and never changes sign.

a) Concave up on \((-\infty, -1)\cup(1, \infty)\); concave down on \((-1, 1)\) b) At \(x=-1\) and \(x=1\), \(f''=0\) at isolated sign-changing zeros. A line has second derivative \(0\) everywhere.

- What must happen to the second derivative at an inflection point? - How are local extrema of \(f'\) related to inflection points of \(f\)? - Does a function change sign at a zero that is also a local minimum of value zero? - Examine the sign of the second derivative on both sides of \(x=1\).

1. Differentiate: \(f'(x)=4x^3-12x^2+12x=4x(x^2-3x+3)\) and \(f''(x)=12x^2-24x+12=12(x-1)^2\). 2. The quadratic \(x^2-3x+3\) has discriminant \(9-12=-3<0\), so the only critical value is \(x=0\). Since \(f''(0)=12>0\), \(f\) has a local minimum at \(x=0\). 3. Although \(f''(1)=0\), the expression \(12(x-1)^2\) is nonnegative on both sides of \(1\), so it does not change sign. 4. Equivalently, \(f''\) decreases to a minimum of \(0\) at \(x=1\) and then increases. Thus, \(f'\) remains increasing and has no local extremum at \(x=1\). Therefore, \(x=1\) is not an inflection value.

a) \(f'(x)=4x^3-12x^2+12x\); \(f''(x)=12x^2-24x+12\) b) A local minimum at \(x=0\) c) No. The second derivative does not change sign at \(x=1\), so \(f'\) has no local extremum there.

- Find every zero of the second derivative. - The condition \(g''(x)=0\) is necessary but not sufficient. - Compare the sign behavior at an even-multiplicity zero and an odd-multiplicity zero. - A sign change in \(g''\) means the graph changes concavity.

1. Differentiate twice: \(g'(x)=\frac{1}{2}x^4-2x^3\) and \(g''(x)=2x^3-6x^2=2x^2(x-3)\). 2. The zeros of \(g''\) are \(x=0\) and \(x=3\). 3. Near \(x=0\), the factor \(2x^2\) is positive on both sides and \(x-3\) is negative. Thus, \(g''(x)<0\) on both sides of \(0\), so there is no inflection point there. 4. At \(x=3\), the factor \(x-3\) changes sign while \(2x^2>0\). Therefore, \(g''\) changes from negative to positive, so \(x=3\) is an inflection value. 5. Evaluate the function: \(g(3)=\frac{243}{10}-\frac{81}{2}+5=-\frac{56}{5}\).

The only inflection point is \(\left(3, -\frac{56}{5}\right)\). The value \(x=0\) is not an inflection value because \(g''\) does not change sign there.

- Which powers can appear in a polynomial that is symmetric about the origin? - Use the necessary and sufficient derivative conditions for an inflection point. - What happens to constants and powers when differentiating?

1. Origin symmetry requires \(f(-x)=-f(x)\), so only odd powers of \(x\) may appear. Therefore, \(b=0\) and \(d=0\). 2. The function becomes \(f(x)=ax^3+cx\). 3. Differentiate: \(f'(x)=3ax^2+c\), \(f''(x)=6ax\), and \(f'''(x)=6a\). 4. Since \(a\ne0\), \(f''(x)=0\) only at \(x=0\), and \(f'''(0)=6a\ne0\). Thus, the graph changes concavity at \(x=0\). 5. Also, \(f(0)=0\), so the inflection point is the origin.

a) \(b=0\) and \(d=0\) b) The graph has an inflection point at \((0, 0)\).

- Which derivative is used to locate inflection points? - When does an equation of the form \(x^2=c\) have two distinct real solutions? - A zero of the second derivative must also produce a sign change. - Use the shape of the second-derivative parabola.

1. Differentiate twice: \(f_k'(x)=2x^3+2kx+7\) and \(f_k''(x)=6x^2+2k\). 2. Solve \(f_k''(x)=0\): \(6x^2+2k=0\), so \(x^2=-\frac{k}{3}\). 3. The equation has two distinct real solutions exactly when \(-\frac{k}{3}>0\), which is equivalent to \(k<0\). 4. For \(k<0\), \(f_k''\) is an upward-opening quadratic with two simple zeros, so it changes sign at both. Thus, the graph has exactly two inflection points.

\(k<0\)

- A graph has no inflection point when the second derivative never changes sign. - When does a quadratic have no crossing of the \(x\)-axis? - Use the discriminant. - What happens to the sign at a double zero?

1. Differentiate twice: \(g_c'(x)=4x^3+6cx^2+12x\) and \(g_c''(x)=12x^2+12cx+12=12(x^2+cx+1)\). 2. The second derivative has no sign-changing zeros when the quadratic \(x^2+cx+1\) has no real zeros or one repeated zero. 3. Its discriminant is \(c^2-4\). Require \(c^2-4\le0\), which gives \(c^2\le4\). 4. Therefore, \(-2\le c\le2\). At the boundary values, the second derivative has a double zero and does not change sign.

\(-2\le c\le2\)

- What derivative condition identifies an inflection point? - Which derivative represents marginal cost? - What happens to the slope when concavity changes? - How are extrema of a derivative related to inflection points of the original function?

1. Differentiate: \(K'(x)=6x^2-72x+300\), \(K''(x)=12x-72\), and \(K'''(x)=12\). 2. Solve \(K''(x)=0\): \(12x-72=0\), so \(x=6\). Since \(K'''(6)=12\ne0\), this is an inflection value. 3. Evaluate the cost: \(K(6)=432-1296+1800+500=1436\). The inflection point is \((6, 1436)\). 4. The marginal-cost function is \(K'\). At \(x=6\), its derivative \(K''\) changes from negative to positive, so \(K'\) has a local minimum. Thus, the cost function's inflection point occurs at the production level where marginal cost is smallest.

a) \((6, 1436)\) b) The inflection value \(x=6\) is where the marginal-cost function \(K'\) reaches a local minimum.

- Which derivative locates inflection points? - How are local extrema of the slope related to concavity changes? - What must the slope equal at a stationary inflection point? - Use the third derivative to classify extrema of the first derivative.

1. Differentiate: \(f'(x)=x^3-3x^2\), \(f''(x)=3x^2-6x\), and \(f'''(x)=6x-6\). 2. Solve \(f''(x)=3x(x-2)=0\): \(x=0\) and \(x=2\). The third derivative is nonzero at both values, so both are inflection values. 3. Since \(f'''(0)=-6<0\), the derivative \(f'\) has a local maximum at \(x=0\). Since \(f'''(2)=6>0\), \(f'\) has a local minimum at \(x=2\). 4. At \(x=0\), \(f(0)=0\) and \(f'(0)=0\), so \((0, 0)\) is a stationary inflection point. 5. At \(x=2\), \(f(2)=-4\) and \(f'(2)=-4\ne0\), so \((2, -4)\) is not stationary.

At \((0, 0)\), the slope has a local maximum and the point is a stationary inflection point. At \((2, -4)\), the slope has a local minimum and the point is not stationary.

- What must the tangent slope equal at a stationary inflection point? - Verify both a horizontal tangent and a change in concavity. - Use the factored second derivative to check the sign change. - Substitute \(x=3\) into the original function.

1. Differentiate once: \(g'(x)=x^3-6x^2+9x\). 2. At \(x=3\), \(g'(3)=27-54+27=0\), so the tangent is horizontal. 3. From \(g''(x)=3(x-1)(x-3)\), \(x=3\) is a simple zero of the second derivative. Therefore, \(g''\) changes sign there, so the graph has an inflection point. 4. The point is therefore a stationary inflection point. 5. Evaluate \(g(3)=\frac{81}{4}-54+\frac{81}{2}-5=\frac{7}{4}\).

Yes. The stationary inflection point is \(\left(3, \frac{7}{4}\right)\).

- Rewrite the rational term using a negative exponent. - Use the second derivative to find an inflection-point candidate. - Check that concavity changes. - A stationary inflection point must also have a horizontal tangent.

1. Rewrite the function as \(f(x)=\frac{1}{2}x^2-4x^{-1}\). 2. Differentiate: \(f'(x)=x+\frac{4}{x^2}\), \(f''(x)=1-\frac{8}{x^3}\), and \(f'''(x)=\frac{24}{x^4}\). 3. Set the second derivative equal to zero: \(1-\frac{8}{x^3}=0\), so \(x^3=8\) and \(x=2\). 4. Since \(f'''(2)=\frac{3}{2}\neq 0\), the concavity changes at \(x=2\). Also, \(f(2)=0\), so the inflection point is \((2, 0)\). 5. Because \(f'(2)=3\neq 0\), the tangent is not horizontal. Therefore, the point is not a stationary inflection point.

The inflection point is \((2, 0)\). It is not a stationary inflection point.

- Start with the general form of a cubic polynomial. - Translate the inflection-point coordinates into conditions on \(f\) and \(f''\). - A horizontal tangent has slope \(0\). - Use the coordinates of each point to form equations.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Then \(f'(x)=3ax^2+2bx+c\) and \(f''(x)=6ax+2b\). 2. Since \(W(0, 2)\) is an inflection point, \(f(0)=2\) and \(f''(0)=0\). Thus, \(d=2\) and \(b=0\). 3. Since \(P(2, 6)\) lies on the graph, \(f(2)=6\), which gives \(4a+c=2\). The tangent at \(x=2\) is horizontal, so \(f'(2)=0\), which gives \(12a+c=0\). 4. Solving the system gives \(a=-\frac{1}{4}\) and \(c=3\). Therefore, \(f(x)=-\frac{1}{4}x^3+3x+2\). Its third derivative is \(-\frac{3}{2}\ne0\), confirming the inflection point.

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

- Start with the general form of a cubic polynomial. - Use the point and inflection conditions at the origin. - A horizontal tangent has slope \(0\). - Solve the resulting system for the remaining coefficients.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Since the origin is on the graph, \(f(0)=0\), so \(d=0\). 2. Since the origin is an inflection point, \(f''(0)=0\). Because \(f''(x)=6ax+2b\), this gives \(b=0\). Thus, \(f(x)=ax^3+cx\). 3. The point \(P(1, 4)\) gives \(a+c=4\). A horizontal tangent at \(P\) gives \(f'(1)=0\), so \(3a+c=0\). 4. Solving the system gives \(a=-2\) and \(c=6\). Therefore, \(f(x)=-2x^3+6x\). Since \(f'''(x)=-12\ne0\), the origin is indeed an inflection point.

\(f(x)=-2x^3+6x\)

- Start with the general cubic form and its first two derivatives. - Translate the y-intercept and slope information into equations. - An inflection point gives a function-value condition and a second-derivative condition. - Solve the resulting system of equations.

1. Let \(f(x)=ax^3+bx^2+cx+d\). The y-intercept gives \(f(0)=3\), so \(d=3\). The slope at the y-intercept gives \(f'(0)=-6\), so \(c=-6\). 2. Since \(W(1, 1)\) is an inflection point, \(f(1)=1\) and \(f''(1)=0\). These conditions give \(a+b=4\) and \(3a+b=0\). 3. Solving the system gives \(a=-2\) and \(b=6\). Therefore, \(f(x)=-2x^3+6x^2-6x+3\). The nonzero third derivative confirms the inflection point.

\(f(x)=-2x^3+6x^2-6x+3\)

- The tangent-line equation gives both the function value and slope at \(x=1\). - Use the second-derivative condition for the inflection point. - Use the origin to determine the constant term. - Solve the resulting linear system.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Since the graph passes through the origin, \(d=0\). 2. At \(x=1\), the tangent line has value \(1\) and slope \(-1\). Therefore, \(f(1)=1\) and \(f'(1)=-1\). The inflection condition gives \(f''(1)=0\). 3. These conditions produce \(a+b+c=1\), \(3a+2b+c=-1\), and \(6a+2b=0\). 4. Solving the system gives \(a=2\), \(b=-6\), and \(c=5\). Therefore, \(f(x)=2x^3-6x^2+5x\). Since \(f'''(x)=12\ne0\), \(x=1\) is indeed an inflection value.

\(f(x)=2x^3-6x^2+5x\)

- Use only even powers for an even quartic polynomial. - The second derivative is \(0\) at an inflection point. - The first derivative gives the tangent slope. - Solve the resulting system for the coefficients.

1. Because the polynomial is even, let \(f(x)=ax^4+bx^2+c\). The point \(S(0, 5)\) gives \(c=5\). 2. The inflection condition \(f''(1)=0\) gives \(12a+2b=0\). The tangent slope condition \(f'(1)=-4\) gives \(4a+2b=-4\). 3. Solving the system gives \(a=\frac{1}{2}\) and \(b=-3\). Therefore, \(f(x)=\frac{1}{2}x^4-3x^2+5\). 4. Since \(f'''(1)=12\ne0\), \(x=1\) is indeed an inflection point.

\(f(x)=\frac{1}{2}x^4-3x^2+5\)

- Translate a stationary inflection point into conditions on \(f\), \(f'\), and \(f''\). - Conditions at the origin simplify the general quartic form. - A horizontal tangent has slope \(0\). - Verify the concavity change with the third derivative.

1. Let \(f(x)=ax^4+bx^3+cx^2+dx+e\). A stationary inflection point at the origin gives \(f(0)=f'(0)=f''(0)=0\), so \(e=d=c=0\). Thus, \(f(x)=ax^4+bx^3\). 2. The point \(P(1, 2)\) gives \(a+b=2\). The horizontal tangent at \(x=\frac{3}{2}\) gives \(f'(\frac{3}{2})=0\), so \(\frac{27}{2}a+\frac{27}{4}b=0\). 3. Solving the system gives \(a=-2\) and \(b=4\). Therefore, \(f(x)=-2x^4+4x^3\). 4. Since \(f'''(0)=24\ne0\), the origin is a stationary inflection point.

\(f(x)=-2x^4+4x^3\)

- Factor the function and retain multiplicity. - Use the second derivative to find the inflection point, then eliminate \(a\). - Use the first derivative to make \(x=4\) critical and the second derivative to classify it.

1. Factor \(f_a(x)=x^2(a-x)\). Thus \(x=0\) is a double zero and \(x=a\) is a zero. 2. The derivatives are \(f_a'(x)=2ax-3x^2\) and \(f_a''(x)=2a-6x\). The second derivative changes sign at \(x=\frac{a}{3}\), so the inflection point has \(x=\frac{a}{3}\) and \(y=\frac{2a^3}{27}\). Since \(a=3x\), substitution gives \(y=2x^3\). 3. For \(x=4\) to be critical, \(f_a'(4)=8a-48=0\), so \(a=6\). Then \(f_6''(4)=12-24=-12<0\), so the point is a local maximum.

1) \(x=0\) is a double zero; \(x=a\) is a zero. 2) \(y=2x^3\) 3) \(a=6\); local maximum

- When does a quadratic have no two distinct real zeros? - Which derivative do you use to locate possible inflection points? - What must the y-coordinate of a point on the x-axis equal? - Factor the equation in \(k\) before solving it.

1. Differentiate: \(g_k'(x)=x^2-2kx+6\). Local extrema require the derivative to change sign at a zero. The quadratic has no two distinct real zeros when its discriminant is nonpositive: \(D=(-2k)^2-4(1)(6)=4k^2-24\le0\). Thus \(k^2\le6\), so \(-\sqrt{6}\le k\le\sqrt{6}\). At the boundary values, the derivative has a double zero but does not change sign, so there is still no local extremum. 2. The second derivative is \(g_k''(x)=2x-2k\). Solving \(g_k''(x)=0\) gives \(x=k\). Since \(g_k'''(x)=2\ne0\), this value gives an inflection point. 3. Evaluate the function at \(x=k\): \(g_k(k)=\frac{1}{3}k^3-k^3+6k=-\frac{2}{3}k^3+6k\). Therefore, \(W_k=\left(k, -\frac{2}{3}k^3+6k\right)\). 4. For the point to lie on the x-axis, solve \(-\frac{2}{3}k^3+6k=0\). Factoring gives \(k\left(-\frac{2}{3}k^2+6\right)=0\), so \(k=-3\), \(k=0\), or \(k=3\).

a) \(-\sqrt{6}\le k\le\sqrt{6}\) b) \(W_k=\left(k, -\frac{2}{3}k^3+6k\right)\) c) \(k\in\{-3, 0, 3\}\)

- Use the inflection point at the origin to determine two coefficients. - The tangent-line slope gives the first derivative at the origin. - Use the additional x-intercept to determine the remaining coefficient. - Check that the third derivative is nonzero.

1. Let \(f(x)=ax^3+bx^2+cx+d\). An inflection point at the origin gives \(f(0)=0\) and \(f''(0)=0\), so \(d=0\) and \(b=0\). 2. The tangent line has slope \(-3\), so \(f'(0)=-3\), giving \(c=-3\). 3. The zero at \(x=3\) gives \(27a-9=0\), so \(a=\frac{1}{3}\). 4. Therefore, \(f(x)=\frac{1}{3}x^3-3x\). Since \(f'''(x)=2\ne0\), the origin is indeed an inflection point.

\(f(x)=\frac{1}{3}x^3-3x\)

- What does the first derivative represent graphically? - To minimize marginal cost, which derivative must equal zero? - Recall the derivative condition for an inflection point. - Interpret the slope of the total-cost curve economically.

1. Marginal cost is \(C'(x)=0.03x^2-1.8x+40\). 2. Differentiate marginal cost: \(C''(x)=0.06x-1.8\). Setting this equal to zero gives \(x=30\). 3. Since \(C'''(x)=0.06>0\), marginal cost has a minimum at \(x=30\). 4. The minimum marginal cost is \(C'(30)=0.03(30)^2-1.8(30)+40=13\), or \(\$13\) per additional unit. 5. Because marginal cost is the slope of \(C\), its minimum occurs where the slope changes from decreasing to increasing. Equivalently, \(C''(30)=0\) and the second derivative changes sign, so \(x=30\) is an inflection point of the total-cost curve.

a) Marginal cost is minimized at \(x=30\) units. b) The minimum marginal cost is \(\$13\) per unit. c) The minimum of marginal cost occurs at the inflection point of \(C\), where the cost curve changes concavity and its slope is smallest.

- Use the second derivative to locate the inflection point. - Verify with the third derivative. - Eliminate the parameter from the coordinates.

1. The derivatives are \(g_k'(x)=x^2+2kx+k^2-1\), \(g_k''(x)=2x+2k\), and \(g_k'''(x)=2\). 2. The inflection point satisfies \(2x+2k=0\), so \(x=-k\). The nonzero third derivative confirms an inflection point. 3. Its y-coordinate is \(g_k(-k)=-\frac{1}{3}k^3+k\). Thus the inflection point is \(\left(-k, -\frac{1}{3}k^3+k\right)\). 4. Since \(k=-x\), substitution gives \(y=\frac{1}{3}x^3-x\).

a) \(\left(-k, -\frac{1}{3}k^3+k\right)\) b) \(y=\frac{1}{3}x^3-x\)

- Isolate the term containing the parameter to find common points. - Apply the critical-point condition at \(x=4\), then use the second derivative. - A stationary inflection point has zero first and second derivatives. - Examine the number of zeros of the second derivative.

1. Rewrite the family as \(f_a(x)=\frac{1}{3}x^3-x+a\left(x-\frac{x^2}{2}\right)\). A point is independent of \(a\) when \(x-\frac{x^2}{2}=0\), so \(x=0\) or \(x=2\). The common points are \((0, 0)\) and \(\left(2, \frac{2}{3}\right)\). 2. Differentiate: \(f_a'(x)=x^2-ax+a-1\). The condition \(f_a'(4)=0\) gives \(16-4a+a-1=0\), so \(a=5\). Since \(f_5''(x)=2x-5\) and \(f_5''(4)=3>0\), the point at \(x=4\) is a local minimum. 3. A stationary inflection point satisfies \(f_a'(x)=0\) and \(f_a''(x)=0\). From \(f_a''(x)=2x-a\), we get \(x=\frac{a}{2}\). Substitution into the first derivative gives \(-\frac{a^2}{4}+a-1=0\), or \((a-2)^2=0\). Thus \(a=2\). Because \(f_a'''(x)=2\neq0\), this critical point is an inflection point. 4. For every \(a\), the equation \(f_a''(x)=2x-a=0\) has exactly one solution, \(x=\frac{a}{2}\). Since the third derivative is always \(2\), the concavity changes there, so each graph has exactly one inflection point.

a) \((0, 0)\) and \(\left(2, \frac{2}{3}\right)\) b) \(a=5\); the extremum at \(x=4\) is a local minimum. c) \(a=2\) d) \(f_a''(x)=2x-a\) has exactly one zero, and \(f_a'''(x)=2\neq0\).

- What must the first and second derivatives equal at a stationary inflection point? - Use those conditions to form a linear system in \(a\) and \(b\). - Check a higher derivative to confirm the change in concavity.

1. A stationary inflection point at \(x=1\) requires \(f'(1)=0\) and \(f''(1)=0\). 2. The derivatives are \(f'(x)=4x^3+3ax^2+2bx\) and \(f''(x)=12x^2+6ax+2b\). 3. Substituting \(x=1\) gives \(3a+2b=-4\) and \(6a+2b=-12\). 4. Subtracting the equations gives \(3a=-8\), so \(a=-\frac{8}{3}\). Substitution gives \(b=2\). 5. The third derivative is \(f'''(x)=24x+6a\), and \(f'''(1)=8\ne0\). Therefore, the concavity changes at \(x=1\), confirming a stationary inflection point.

\(a=-\frac{8}{3}\), \(b=2\)

- A local extremum must satisfy a first-derivative condition. - Use the second derivative to classify it. - Count the roots of the second derivative and check whether its sign changes.

1. The first derivative is \(f_k'(x)=x^3+kx+2\). Requiring \(f_k'(-1)=0\) gives \(-1-k+2=0\), so \(k=1\). 2. The second derivative is \(f_k''(x)=3x^2+k\). Since \(f_1''(-1)=4>0\), the point is a local minimum. 3. Inflection points require a sign change in \(f_k''(x)=3x^2+k\). If \(k<0\), the equation \(3x^2+k=0\) has two distinct roots, and the second derivative changes sign at both, so there are two inflection points. 4. If \(k>0\), the second derivative is always positive, so there are no inflection points. If \(k=0\), its only zero is \(x=0\), but there is no sign change. Therefore, there are no inflection points for \(k\ge0\).

a) \(k=1\); local minimum b) \(k<0\): two inflection points; \(k\ge0\): no inflection points

- Check whether the denominator can be zero and compare \(f(-x)\) with \(f(x)\). - Compare the numerator and denominator degrees for end behavior. - Use the first derivative for extrema and the second derivative for inflection points. - Verify sign changes at all derivative zeros.

1. Since \(x^2+4>0\) for every real \(x\), the domain is \(\mathbb{R}\). 2. Since \(f(-x)=-f(x)\), the graph is symmetric about the origin. The x- and y-intercepts are both \((0, 0)\). 3. The denominator has greater degree than the numerator, so \(\lim_{x\to\pm\infty}f(x)=0\). The horizontal asymptote is \(y=0\). 4. The first derivative is \(f'(x)=\frac{16-4x^2}{(x^2+4)^2}\). It is zero at \(x=\pm 2\). The derivative changes from negative to positive at \(x=-2\), giving a local minimum at \((-2, -1)\), and from positive to negative at \(x=2\), giving a local maximum at \((2, 1)\). 5. The second derivative is \(f''(x)=\frac{8x(x^2-12)}{(x^2+4)^3}\). Its zeros are \(x=-2\sqrt{3}\), \(x=0\), and \(x=2\sqrt{3}\), and its sign changes at each one. 6. The inflection points are \(\left(-2\sqrt{3}, -\frac{\sqrt{3}}{2}\right)\), \((0, 0)\), and \(\left(2\sqrt{3}, \frac{\sqrt{3}}{2}\right)\).

Domain: \(\mathbb{R}\); symmetric about the origin; intercept: \((0, 0)\); horizontal asymptote: \(y=0\); local minimum: \((-2, -1)\); local maximum: \((2, 1)\); inflection points: \(\left(-2\sqrt{3}, -\frac{\sqrt{3}}{2}\right)\), \((0, 0)\), and \(\left(2\sqrt{3}, \frac{\sqrt{3}}{2}\right)\)

- Use the chain rule for the first derivative and then differentiate again. - An inflection point requires a change in concavity. - Factor \(x^2\) from under the square root when evaluating the limits. - To find each asymptote, evaluate the limit of the difference between \(f(x)\) and a line with the corresponding slope.

1. Differentiate: \(f'(x)=\frac{x}{\sqrt{x^2+5}}\) and \(f''(x)=\frac{5}{(x^2+5)^{3/2}}\). 2. Solve \(f'(x)=0\). This gives \(x=0\). Since \(f''(0)=\frac{1}{\sqrt{5}}>0\), the point is a local minimum. Its coordinates are \((0,\sqrt{5})\). 3. Because \(f''(x)>0\) for every real \(x\), the concavity never changes. Therefore, the graph has no inflection points. 4. Using \(\sqrt{x^2}=|x|\), \(\lim_{x\to\infty}\frac{x}{\sqrt{x^2+5}}=1\) and \(\lim_{x\to-\infty}\frac{x}{\sqrt{x^2+5}}=-1\). 5. For the right-hand asymptote, \(\lim_{x\to\infty}(f(x)-x)=\lim_{x\to\infty}\frac{5}{\sqrt{x^2+5}+x}=0\), so the asymptote is \(y=x\). For the left-hand asymptote, \(\lim_{x\to-\infty}(f(x)+x)=\lim_{x\to-\infty}\frac{5}{\sqrt{x^2+5}-x}=0\), so the asymptote is \(y=-x\). Their slopes are the derivative limits from part c).

a) Local minimum at \((0,\sqrt{5})\). b) Since \(f''(x)=\frac{5}{(x^2+5)^{3/2}}>0\) for all real \(x\), there are no inflection points. c) \(\lim_{x\to\infty}f'(x)=1\) and \(\lim_{x\to-\infty}f'(x)=-1\). d) The slant asymptotes are \(y=x\) as \(x\to\infty\) and \(y=-x\) as \(x\to-\infty\). Their slopes are \(1\) and \(-1\), respectively.

- Count the real roots of the first derivative using its discriminant. - Set the second derivative equal to zero. - Substitute the inflection x-coordinate into the original function.

1. The first three derivatives are \(g_a'(x)=x^2+2ax+4\), \(g_a''(x)=2x+2a\), and \(g_a'''(x)=2\). 2. Critical points satisfy \(x^2+2ax+4=0\). This quadratic has two distinct real roots when its discriminant is positive: \(4a^2-16>0\), or \(|a|>2\). At those roots, the second derivative is nonzero, so both are local extrema. 3. An inflection point occurs where \(g_a''(x)=0\). Thus \(2x+2a=0\), so \(x=-a\). Because \(g_a'''(x)=2\ne0\), this is an inflection point. 4. Its y-coordinate is \(g_a(-a)=-\frac{1}{3}a^3+a^3-4a=\frac{2}{3}a^3-4a\).

a) \(a<-2\) or \(a>2\) b) \(\left(-a, \frac{2}{3}a^3-4a\right)\)

- Set the second derivative equal to zero. - Use the fact that \(e^x\) is always positive. - Verify that the second derivative changes sign. - Substitute the resulting exponential value directly into the original function.

1. The second derivative is \(f_a''(x)=e^x-4ae^{2x}=e^x(1-4ae^x)\). 2. Since \(e^x>0\), the equation \(f_a''(x)=0\) is equivalent to \(e^x=\frac{1}{4a}\). This has a real solution exactly when \(a>0\). The factor \(1-4ae^x\) changes sign at that solution, so it is an inflection point. 3. Its x-coordinate is \(x=-\ln(4a)\). 4. Substituting \(e^x=\frac{1}{4a}\) gives \(f_a(x)=\frac{1}{4a}-a\left(\frac{1}{4a}\right)^2=\frac{3}{16a}\).

1. An inflection point exists exactly when \(a>0\). 2. \(\left(-\ln(4a), \frac{3}{16a}\right)\)

- Use the product rule and chain rule for successive derivatives. - Set the second derivative equal to zero. - Verify that the candidate is an inflection point. - Compare the x- and y-coordinates of the inflection point.

1. By the product and chain rules, \(f_t'(x)=(1+tx)e^{tx}\) and \(f_t''(x)=(2t+t^2x)e^{tx}\). 2. Since \(e^{tx}>0\), the inflection-point equation gives \(2t+t^2x=0\), so \(x=-\frac{2}{t}\). 3. The third derivative is \(f_t'''(x)=(3t^2+t^3x)e^{tx}\). At \(x=-\frac{2}{t}\), it equals \(t^2e^{-2}>0\), confirming an inflection point. 4. The y-coordinate is \(f_t\left(-\frac{2}{t}\right)=-\frac{2}{te^2}\). Thus the inflection point is \(\left(-\frac{2}{t}, -\frac{2}{te^2}\right)\). 5. Since \(y=\frac{1}{e^2}x\), all inflection points lie on the line \(y=\frac{x}{e^2}\).

a) \(\left(-\frac{2}{t}, -\frac{2}{te^2}\right)\) b) \(y=\frac{x}{e^2}\)

- The argument of a logarithm must be positive. - Use the first derivative to locate critical numbers and the second derivative to classify them. - A constant sign for the second derivative means the concavity does not change. - Examine the function as \(x\) approaches the excluded value.

1. The logarithm requires \(x^2>0\), so the domain is \(\mathbb{R}\setminus\{0\}\). 2. Differentiate: \(f'(x)=\frac{2}{x}-x+2\) and \(f''(x)=-\frac{2}{x^2}-1\). 3. Solve \(f'(x)=0\): \(\frac{2-x^2+2x}{x}=0\), so \(x^2-2x-2=0\). The critical numbers are \(x=1\pm\sqrt{3}\). 4. Since \(f''(x)<0\) everywhere in the domain, both critical points are local maxima. Using \(x^2=2x+2\) at either critical number, their coordinates are \(\left(1+\sqrt{3},\ln(4+2\sqrt{3})+\sqrt{3}\right)\) and \(\left(1-\sqrt{3},\ln(4-2\sqrt{3})-\sqrt{3}\right)\). 5. The second derivative is never zero and is negative on both domain intervals. Thus the graph is concave down throughout its domain and has no inflection points. 6. The point \(x=0\) is not in the domain, and \(f(x)\to-\infty\) as \(x\to0^-\) or \(x\to0^+\). The vertical asymptote separates the graph into two branches, so the two maxima are not connected by a continuous portion of the graph.

a) \(\mathbb{R}\setminus\{0\}\). b) Local maxima at \(\left(1+\sqrt{3},\ln(4+2\sqrt{3})+\sqrt{3}\right)\) and \(\left(1-\sqrt{3},\ln(4-2\sqrt{3})-\sqrt{3}\right)\). c) There are no inflection points. d) The vertical asymptote \(x=0\) separates the graph into two branches, so no connected part of the graph must pass through a minimum or an inflection point between the maxima.

- Compare \(f_k(-x)\) with \(f_k(x)\). - A natural logarithm is zero when its argument is \(1\). - Use the quotient rule for the second derivative. - Tangency to the x-axis requires both a zero function value and zero slope.

1. Since \(f_k(-x)=f_k(x)\), each graph is symmetric about the y-axis. Also, \(x^2+k\to\infty\) as \(x\to\pm\infty\), so \(f_k(x)\to\infty\). 2. Zeros satisfy \(x^2+k=1\), or \(x^2=1-k\). Thus there are two zeros when \(0<k<1\), one zero at \(x=0\) when \(k=1\), and no zeros when \(k>1\). 3. The second derivative is \(f_k''(x)=\frac{2(k-x^2)}{(x^2+k)^2}\). It is zero at \(x=\pm\sqrt{k}\), and its sign changes at both points. Therefore, there are exactly two inflection points. 4. Tangency at the origin requires \(f_k(0)=0\) and \(f_k'(0)=0\). The first condition gives \(\ln(k)=0\), so \(k=1\); the derivative condition is then also satisfied.

1. Symmetric about the y-axis; \(f_k(x)\to\infty\) as \(x\to\pm\infty\) 2. \(0<k<1\): two zeros; \(k=1\): one zero; \(k>1\): no zeros 3. \(x=\pm\sqrt{k}\) 4. \(k=1\)

- The argument of a logarithm must be positive. - Examine the function as the argument approaches \(0^+\). - Find the critical point from the first derivative. - Study the sign of the second derivative on the entire domain.

1. The logarithm requires \(a-x^2>0\), so the maximal domain is \((-\sqrt{a},\sqrt{a})\). As \(x\to(-\sqrt{a})^{+}\) or \(x\to(\sqrt{a})^{-}\), the argument approaches \(0^+\), so \(g_a(x)\to-\infty\). Thus, \(x=\pm\sqrt{a}\) are vertical asymptotes. 2. The first derivative is \(g_a'(x)=\frac{-2x}{a-x^2}\), which is zero only at \(x=0\). 3. The second derivative is \(g_a''(x)=-\frac{2(a+x^2)}{(a-x^2)^2}\), which is negative throughout the domain. Therefore, \((0, \ln(a))\) is a local and absolute maximum. 4. Because the second derivative is always negative and never zero, the graph has no inflection points.

1. Domain: \((-\sqrt{a},\sqrt{a})\); \(g_a(x)\to-\infty\) at both endpoints 2. Maximum: \((0, \ln(a))\) 3. No inflection points

- The exponential factor is defined and positive for every real input. - Compare exponential and polynomial growth for large negative values of \(x\). - Apply the product rule and factor out \(e^x\) after differentiating. - Use the first and second derivatives to locate and classify the requested points.

1. The domain is all real numbers. Since \(e^x\ne0\), the x-intercepts satisfy \(x^2-3=0\), so they are \((-\sqrt{3},0)\) and \((\sqrt{3},0)\). The y-intercept is \((0,-3)\). 2. As \(x\to\infty\), both factors grow without bound, so \(f(x)\to\infty\). As \(x\to-\infty\), exponential decay dominates polynomial growth, so \(f(x)\to0\). 3. The first derivative is \(f'(x)=(x^2+2x-3)e^x=(x-1)(x+3)e^x\). Thus the critical numbers are \(x=-3\) and \(x=1\). The second derivative is \(f''(x)=(x^2+4x-1)e^x\). Since \(f''(-3)=-4e^{-3}<0\), there is a local maximum at \((-3,6e^{-3})\). Since \(f''(1)=4e>0\), there is a local minimum at \((1,-2e)\). 4. Inflection points can occur where \(f''(x)=0\). Solving \(x^2+4x-1=0\) gives \(x=-2\pm\sqrt{5}\). Both roots are simple, so the concavity changes at each one.

1) Domain: \(\mathbb{R}\). x-intercepts: \((-\sqrt{3},0)\), \((\sqrt{3},0)\). y-intercept: \((0,-3)\). 2) \(\lim_{x\to\infty}f(x)=\infty\) and \(\lim_{x\to-\infty}f(x)=0\). 3) Local maximum at \((-3,6e^{-3})\); local minimum at \((1,-2e)\). 4) The inflection-point x-coordinates are \(-2-\sqrt{5}\) and \(-2+\sqrt{5}\).

- Find the inflection points using the second derivative. - Use the symmetry of the graph about the y-axis. - Two vectors are perpendicular when their dot product is zero.

1. Differentiate twice: \(f_k'(x)=-2kxe^{-kx^2}\) and \(f_k''(x)=(4k^2x^2-2k)e^{-kx^2}\). 2. Since the exponential factor is positive, inflection-point candidates satisfy \(4k^2x^2-2k=0\), giving \(x=\pm\frac{1}{\sqrt{2k}}\). The second derivative changes sign at both values, so they are inflection points. 3. Their y-coordinate is \(e^{-1/2}=\frac{1}{\sqrt{e}}\). Thus the inflection points are \(\left(\frac{1}{\sqrt{2k}}, \frac{1}{\sqrt{e}}\right)\) and \(\left(-\frac{1}{\sqrt{2k}}, \frac{1}{\sqrt{e}}\right)\). 4. The angle at the origin is right when the position vectors to the two inflection points have dot product zero. Therefore, \(-\frac{1}{2k}+\frac{1}{e}=0\). 5. Solving gives \(2k=e\), so \(k=\frac{e}{2}\).

\(k=\frac{e}{2}\)

- Find the inflection points in terms of \(k\). - Use their equal y-coordinates to choose a horizontal base. - Write the triangle’s area before solving for \(k\).

1. Differentiate twice: \(f_k'(x)=-2kxe^{-kx^2}\) and \(f_k''(x)=2k(2kx^2-1)e^{-kx^2}\). 2. Since the exponential factor is positive, the candidates have \(x=\pm\frac{1}{\sqrt{2k}}\). The factor \(2kx^2-1\) changes sign at both values, so they are inflection points. Their common y-coordinate is \(e^{-1/2}=\frac{1}{\sqrt{e}}\). 3. The triangle has base \(\frac{2}{\sqrt{2k}}\) and height \(\frac{1}{\sqrt{e}}\). Thus \(A=\frac{1}{2}\cdot\frac{2}{\sqrt{2k}}\cdot\frac{1}{\sqrt{e}}=\frac{1}{\sqrt{2ke}}\). 4. Setting \(\frac{1}{\sqrt{2ke}}=\frac{1}{\sqrt{2e}}\) gives \(k=1\).

\(k=1\)

- Set the second derivative equal to zero. - Verify the inflection point using the third derivative. - Eliminate \(a\) from the point coordinates. - Include the restriction implied by \(a\neq0\).

1. Differentiate: \(f_a''(x)=6x-12a\). The inflection-point equation gives \(x=2a\). 2. Since \(f_a'''(x)=6\neq0\), this point is an inflection point. 3. Its y-coordinate is \(f_a(2a)=8a^3-24a^3+2=-16a^3+2\). 4. From \(x=2a\), obtain \(a=\frac{x}{2}\). Substitution gives \(y=-16\left(\frac{x}{2}\right)^3+2=-2x^3+2\). 5. Because \(a\neq0\), the attainable x-values satisfy \(x\neq0\).

\(y=-2x^3+2\) for \(x\neq0\)

- Rewrite the function using negative exponents before differentiating. - Which derivative must change sign at an inflection point? - Check that the candidate value is in the domain for every \(k>0\). - What must the first derivative equal at a stationary inflection point?

1. Rewrite and differentiate: \(f_k(x)=x+3x^{-1}+kx^{-2}\), \(f_k'(x)=1-3x^{-2}-2kx^{-3}\), and \(f_k''(x)=6x^{-3}+6kx^{-4}=\frac{6(x+k)}{x^4}\). 2. Since \(x^4>0\) for every \(x\ne0\), the sign of \(f_k''(x)\) is determined by \(x+k\). The second derivative has exactly one zero, \(x=-k\), and changes sign there. Because \(k>0\), \(-k\ne0\), so this value is in the domain. Thus every function has exactly one inflection point. 3. Its y-coordinate is \(f_k(-k)=-k-\frac{3}{k}+\frac{1}{k}=-k-\frac{2}{k}\), so the point is \(\left(-k, -k-\frac{2}{k}\right)\). 4. The inflection point is stationary when \(f_k'(-k)=0\): \(f_k'(-k)=1-\frac{3}{k^2}+\frac{2}{k^2}=1-\frac{1}{k^2}\). Therefore, \(k^2=1\). Since \(k>0\), \(k=1\).

a) Every function has exactly one inflection point, \(\left(-k, -k-\frac{2}{k}\right)\). b) The inflection point is stationary when \(k=1\).

- Look for where the graph changes from bending downward to bending upward. - Read both coordinates of the concavity-change point from the grid. - Which derivative gives the slope of the tangent line? - Use the tangent slope and the point \(W\) in point-slope form.

1. The graph changes concavity at \(x=3\). It is concave down for \(x<3\) and concave up for \(x>3\). 2. The inflection point is \(W=(3, 2)\). 3. Differentiate: \(f'(x)=\frac{3}{4}(x-3)^2-3\). Therefore, the tangent slope is \(f'(3)=-3\). 4. Using point-slope form through \((3, 2)\), \(y-2=-3(x-3)\), so the tangent line is \(y=-3x+11\).

a) Concave down for \(x<3\); concave up for \(x>3\). b) \(W=(3, 2)\) c) \(y=-3x+11\)

- Relate zeros and sign changes of \(f'\) to local extrema of \(f\). - Use the sign of \(f'\) to determine where \(f\) increases or decreases. - Inflection points of \(f\) correspond to local extrema of \(f'\), not merely zeros of \(f'\).

1. The graph of \(f'\) crosses the \(x\)-axis at \(x = -2\), \(x = 0\), and \(x = 3\). Each zero has a sign change, so \(f\) has exactly three local extrema in \((-2.5, 3.5)\). Statement a) is true. 2. The graph of \(f'\) is below the \(x\)-axis throughout \([0.5, 2.5]\), so \(f'(x) < 0\) there and \(f\) is strictly decreasing. Statement b) is true. 3. An inflection point of \(f\) occurs where \(f'\) has a local extremum, because that is where \(f''\) changes sign. At \(x = 0\), \(f'\) crosses the axis but does not have a local extremum. Statement c) is false.

a) True b) True c) False

- Set the first derivative equal to zero to find horizontal tangents. - Use the second and third derivatives to test whether concavity changes. - A stationary inflection point is an inflection point with slope zero. - Solve \(f''(x)=0\) to find all possible inflection points.

1. Differentiate: \(f'(x)=60x^4-120x^3+60x^2=60x^2(x-1)^2\). 2. Solving \(f'(x)=0\) gives the critical values \(x=0\) and \(x=1\). 3. The next derivatives are \(f''(x)=120x(2x-1)(x-1)\) and \(f'''(x)=720x^2-720x+120\). 4. At \(x=0\), \(f''(0)=0\), \(f'''(0)=120\ne0\), and \(f'(0)=0\). Thus, \((0,0)\) is a stationary inflection point. At \(x=1\), \(f''(1)=0\), \(f'''(1)=120\ne0\), and \(f'(1)=0\). Thus, \((1,2)\) is also a stationary inflection point. 5. Solving \(f''(x)=0\) gives \(x=0\), \(x=\frac{1}{2}\), and \(x=1\). At \(x=\frac{1}{2}\), \(f\left(\frac{1}{2}\right)=1\) and \(f'\left(\frac{1}{2}\right)=\frac{15}{4}=3.75\). Therefore, the other inflection point is \(\left(\frac{1}{2},1\right)\). Its slope is not zero, so it is not a stationary inflection point.

a) \(x=0\) and \(x=1\) b) Stationary inflection points: \((0,0)\) and \((1,2)\) c) Inflection point: \(\left(\frac{1}{2},1\right)\); slope: \(\frac{15}{4}=3.75\). It is not stationary because the slope is not zero.

- Look for where the graph changes how it bends. - Which derivative determines concavity? - What equation gives possible inflection values? - Use the sign of the second derivative on each interval. - Substitute the inflection values into the original function.

1. From the graph, the concavity appears to change near \(x=-1\) and \(x=1\). The graph appears concave up between these values and concave down outside them. 2. Differentiate: \(f'(x)=-0.5x^3+1.5x\), \(f''(x)=-1.5x^2+1.5\), and \(f'''(x)=-3x\). 3. Solve \(f''(x)=0\): \(-1.5x^2+1.5=0\), so \(x=\pm1\). Since \(f'''(-1)=3\ne0\) and \(f'''(1)=-3\ne0\), both values give inflection points. 4. Evaluate the function: \(f(-1)=f(1)=\frac{5}{8}\). Thus the inflection points are \(\left(-1, \frac{5}{8}\right)\) and \(\left(1, \frac{5}{8}\right)\). 5. The second derivative is positive on \((-1, 1)\), so \(f\) is concave up there. It is negative on \((-\infty, -1)\) and \((1, \infty)\), so \(f\) is concave down there.

a) Approximately concave up on \((-1, 1)\) and concave down outside that interval; the inflection points are approximately \((-1, 0.6)\) and \((1, 0.6)\). b) The exact inflection points are \(\left(-1, \frac{5}{8}\right)\) and \(\left(1, \frac{5}{8}\right)\). The graph is concave up on \((-1, 1)\) and concave down on \((-\infty, -1)\) and \((1, \infty)\).

- Identify the local maximum and local minimum on the graph. - For a cubic polynomial, how is the inflection point positioned between the two extrema? - Where does the graph change from bending downward to bending upward? - Read the inflection point’s coordinates from the grid.

1. The local maximum \((1, 3)\) and local minimum \((3, -1)\) are symmetric about the cubic graph’s inflection point. Their midpoint is \(W=\left(\frac{1+3}{2}, \frac{3+(-1)}{2}\right)=(2, 1)\). 2. In the displayed window, the graph bends downward for \(0\le x<2\), so it is concave down on \([0, 2)\). 3. The graph bends upward for \(2<x\le4\), so it is concave up on \((2, 4]\).

a) \([0, 2)\) b) \((2, 4]\) c) \(W=(2, 1)\)

- Locate where the graph appears to change how it bends. - Which derivative determines concavity? - Solve the equation obtained by setting the second derivative equal to zero. - Use test values to determine the sign of the second derivative on each interval.

1. Differentiate: \(f'(x)=0.4x^3-1.2x+0.4\), \(f''(x)=1.2x^2-1.2\), and \(f'''(x)=2.4x\). 2. Solve \(f''(x)=0\): \(1.2x^2-1.2=0\), so \(x=-1\) or \(x=1\). Since \(f'''(-1)=-2.4\ne0\) and \(f'''(1)=2.4\ne0\), both values give inflection points. 3. Evaluate the function: \(f(-1)=-0.9\) and \(f(1)=-0.1\). Therefore, the inflection points are \((-1, -0.9)\) and \((1, -0.1)\). 4. The second derivative is positive for \(x<-1\) and \(x>1\), so the graph is concave up on \((-\infty, -1)\) and \((1, \infty)\). It is negative for \(-1<x<1\), so the graph is concave down on \((-1, 1)\).

a) \(x\approx-1\) and \(x\approx1\) b) \((-1, -0.9)\) and \((1, -0.1)\) c) Concave up on \((-\infty, -1)\) and \((1, \infty)\); concave down on \((-1, 1)\).

- Which derivative determines concavity? - What equation gives possible inflection values? - How can the third derivative verify an inflection point? - Use the sign of the second derivative on each interval.

1. Differentiate: \(f'(x)=0.4x^3-1.2x\), \(f''(x)=1.2x^2-1.2\), and \(f'''(x)=2.4x\). 2. Solve \(f''(x)=0\): \(1.2x^2-1.2=0\), so \(x=-1\) or \(x=1\). Since \(f'''(-1)=-2.4\ne0\) and \(f'''(1)=2.4\ne0\), both values give inflection points. 3. Evaluate the function: \(f(-1)=0\) and \(f(1)=0\). Therefore, the inflection points are \((-1, 0)\) and \((1, 0)\). 4. The second derivative is positive when \(x<-1\) or \(x>1\), so the graph is concave up on \((-\infty, -1)\) and \((1, \infty)\). It is negative when \(-1<x<1\), so the graph is concave down on \((-1, 1)\). These changes match the graph.

1) \((-1, 0)\) and \((1, 0)\) 2) Concave up on \((-\infty, -1)\) and \((1, \infty)\); concave down on \((-1, 1)\).

- Look for where the graph changes how it bends. - Read the coordinates of that change from the grid. - What sign does the second derivative have where a graph is concave down or concave up? - Locate \(x=-1\) and \(x=3\) before judging the concavity there.

1. On the displayed domain \([-3.4, 5.3]\), the graph changes concavity at \(x=1\). It is concave down on \([-3.4, 1)\) and concave up on \((1, 5.3]\). 2. Reading from the graph gives the inflection point \(W\approx(1, 0.9)\). 3. At \(x=-1\), the graph is concave down, so \(f''(-1)<0\). At \(x=3\), the graph is concave up, so \(f''(3)>0\).

a) Concave down on \([-3.4, 1)\); concave up on \((1, 5.3]\). b) \(W\approx(1, 0.9)\) c) \(f''(-1)<0\) and \(f''(3)>0\)

- An inflection point is where the graph changes concavity. - How does the slope change on a concave-up interval? On a concave-down interval? - A change from increasing slopes to decreasing slopes produces what kind of extremum of \(f'\)? - Use the displayed domain when writing the intervals.

1. The graph changes concavity at \(x=1\) and \(x=3\), so these are the inflection values. 2. At \(x=1\), the graph changes from concave up to concave down. Therefore, the slope changes from increasing to decreasing, so \(f'\) has a local maximum there. In fact, \(f'(1)=8\). 3. At \(x=3\), the graph changes from concave down to concave up. Therefore, the slope changes from decreasing to increasing, so \(f'\) has a local minimum there. In fact, \(f'(3)=-8\). 4. On the displayed domain, \(f\) is concave up on \([-0.5, 1)\) and \((3, 4.5]\), and concave down on \((1, 3)\).

a) \(x=1\) and \(x=3\) b) \(f'\) has a local maximum at \(x=1\) and a local minimum at \(x=3\). c) Concave up on \([-0.5, 1)\) and \((3, 4.5]\); concave down on \((1, 3)\).

- Use the sign of the first derivative to determine increasing and decreasing intervals. - Use the sign of the second derivative to determine concavity. - A point where the second derivative is \(0\) is not part of an interval where it is strictly positive or negative.

1. The derivative is \(f'(x)=\frac{3}{4}x^2-3=\frac{3}{4}(x-2)(x+2)\). It is positive for \(x<-2\) and \(x>2\), and negative for \(-2<x<2\). Therefore, \(f\) is strictly increasing on \((-\infty, -2]\) and \([2, \infty)\), and strictly decreasing on \([-2, 2]\). 2. The second derivative is \(f''(x)=\frac{3}{2}x\). It is negative for \(x<0\) and positive for \(x>0\). Therefore, the graph is concave down on \((-\infty, 0)\) and concave up on \((0, \infty)\).

a) Increasing on \((-\infty, -2]\) and \([2, \infty)\); decreasing on \([-2, 2]\) b) Concave down on \((-\infty, 0)\); concave up on \((0, \infty)\)

- Compare the graph’s right-end behavior with each exponential factor. - Use the graph’s x-intercept to eliminate choices. - A local maximum has a horizontal tangent. - Concavity is determined by the sign of the second derivative. - Differentiate the selected function to verify both statements.

1. The graph approaches the x-axis as \(x\to\infty\), so option A is not possible because \((x+2)e^x\to\infty\). 2. The graph has an x-intercept at \(x=-2\), which rules out option B. Therefore, option C is the matching function: \(f(x)=(x+2)e^{-x}\). 3. Differentiate: \(f'(x)=-(x+1)e^{-x}\). Thus \(f'(-1)=0\), so statement (1) is true. This also agrees with the local maximum shown at \(x=-1\). 4. The second derivative is \(f''(x)=xe^{-x}\). For \(x<0\), the exponential factor is positive and \(x\) is negative, so \(f''(x)<0\). The graph is concave down, not concave up, so statement (2) is false.

a) (C) \(f_3(x)=(x+2)e^{-x}\). b) (1) True. \(f'(-1)=0\). (2) False. Since \(f''(x)=xe^{-x}<0\) for \(x<0\), the graph is concave down there.

- Rewrite the expression as a quotient to analyze its end behavior. - Apply the product and chain rules. - Use the first and second derivatives to locate and classify the local maximum. - Verify a change in concavity at the candidate inflection point. - The exponential factor is never zero.

1. Rewrite \(f(x)=\frac{x+1}{e^x}\). Exponential growth dominates linear growth, so \(\lim_{x\to\infty}f(x)=0\). Equivalently, L’Hôpital’s rule gives \(\lim_{x\to\infty}\frac{1}{e^x}=0\). 2. Differentiate: \(f'(x)=-xe^{-x}\). The only critical number is \(x=0\). Since \(f''(x)=(x-1)e^{-x}\) and \(f''(0)=-1<0\), the graph has a local maximum at \((0,1)\). 3. The second derivative is zero at \(x=1\). Since \(f'''(x)=(2-x)e^{-x}\) and \(f'''(1)=e^{-1}\ne0\), the graph has an inflection point at \(\left(1,\frac{2}{e}\right)\). 4. Since \(e^{-x}\ne0\), \((x+1)e^{-x}=0\) only when \(x=-1\).

a) \(\lim_{x\to\infty}f(x)=0\). b) \((0,1)\). c) \(\left(1,\frac{2}{e}\right)\). d) \(x=-1\).

- A maximum occurs at a critical point of the concentration function. - In part b, set the function equal to the given concentration. - The equation in part b has two solutions; choose the one after the maximum. - Interpret the slope of the graph as the rate at which concentration changes.

1. Differentiate: \(C'(t) = \frac{10 - 10t^2}{(t^2 + 1)^2}\). 2. Solve \(C'(t) = 0\). Since \(t \ge 0\), the relevant critical time is \(t = 1\). Also, \(C''(t) = \frac{20t(t^2 - 3)}{(t^2 + 1)^3}\), and \(C''(1) = -5 < 0\), so the concentration is maximized at \(t = 1\). The maximum is \(C(1) = 5\,\text{mg/L}\). 3. Solve \(\frac{10t}{t^2 + 1} = 4\). This gives \(2t^2 - 5t + 2 = 0\), with solutions \(t = \frac{1}{2}\) and \(t = 2\). The time after the maximum is \(t = 2\,\text{h}\). 4. At \(t = \sqrt{3}\), the concavity changes and \(C'(t)\) reaches its minimum value. In context, the concentration is decreasing at its fastest rate at that time.

a) The maximum concentration is \(5\,\text{mg/L}\) at \(t = 1\,\text{h}\). b) After the maximum, the concentration is \(4\,\text{mg/L}\) at \(t = 2\,\text{h}\). c) At \(t = \sqrt{3}\,\text{h}\), the concentration is decreasing most rapidly.

- Set the second derivative equal to zero, then use the third derivative to verify each candidate. - Substitute each verified x-value into the original function.

1. For part a, differentiate: \(f'(x)=x^3-6x^2+9x\), \(f''(x)=3x^2-12x+9\), and \(f'''(x)=6x-12\). 2. Solve \(f''(x)=3(x-1)(x-3)=0\), giving \(x=1\) and \(x=3\). Since \(f'''(1)=-6\ne0\) and \(f'''(3)=6\ne0\), both are inflection values. Also, \(f(1)=-\frac{9}{4}\) and \(f(3)=\frac{7}{4}\). 3. For part b, differentiate: \(g'(x)=\frac{1}{2}x^2-2x+1\), \(g''(x)=x-2\), and \(g'''(x)=1\). 4. Solving \(g''(x)=0\) gives \(x=2\). Since \(g'''(2)=1\ne0\), this is an inflection value. Evaluating gives \(g(2)=\frac{10}{3}\).

a) \(\left(1, -\frac{9}{4}\right)\) and \(\left(3, \frac{7}{4}\right)\) b) \(\left(2, \frac{10}{3}\right)\)

- Differentiate each function three times. - Zeros of the second derivative are candidates for inflection values. - A nonzero third derivative verifies a candidate in these problems.

1. For part a, \(f'(x)=4x^3-12x\), \(f''(x)=12x^2-12\), and \(f'''(x)=24x\). Solving \(f''(x)=0\) gives \(x=-1\) and \(x=1\). Since \(f'''(-1)=-24\ne0\) and \(f'''(1)=24\ne0\), both are inflection values. Also, \(f(-1)=f(1)=3\). 2. For part b, \(g'(x)=6x^2-24x+18\), \(g''(x)=12x-24\), and \(g'''(x)=12\). Solving \(g''(x)=0\) gives \(x=2\). Since \(g'''(2)=12\ne0\), this is an inflection value. Also, \(g(2)=4\).

a) \((-1, 3)\) and \((1, 3)\) b) \((2, 4)\)

- An extremum of \(f'\) can correspond to an inflection point of \(f\). - Use the sign of \(f'\) between \(x=1\) and \(x=3\). - The tangent slope of \(f\) at a point equals the value of \(f'\) there.

1. The graph of \(f'\) has a local maximum at \(x=3\). Its slope, \(f''\), changes from positive to negative there, so \(f\) changes concavity. Therefore, \(f\) has an inflection point at \(x=3\). 2. The derivative satisfies \(f'(x)>0\) for \(1<x<3\), so \(f\) is strictly increasing on \([1,3]\). Therefore, \(f(3)>f(1)\). 3. The tangent slope at \(x=0\) is \(f'(0)\). From the graph, \(f'(0)=-2.5\).

a) \(x=3\), because \(f'\) has a local maximum there and \(f''\) changes sign b) \(f(3)>f(1)\) c) \(f'(0)=-2.5\)

- Decide which graph can be the derivative of the other. - A function is concave up where its second derivative is positive.

1. The red graph is cubic, and the blue graph is its quadratic derivative. Therefore, \(g_1 = f'\) and \(g_2 = f''\). 2. The graph of \(f\) is concave up where \(f''(x) > 0\). The blue graph is above the \(x\)-axis between its zeros \(x = -2\) and \(x = 2\). 3. Therefore, \(f\) is concave up on \((-2, 2)\).

a) \(g_1 = f'\) and \(g_2 = f''\) b) \((-2, 2)\)

- Use the second derivative to locate the inflection value. - Substitute the x-value into the original function. - Use the first derivative to find the tangent slope. - Write the tangent line in point-slope form.

1. Differentiate: \(f'(x)=\frac{1}{2}x^2-x-1\), \(f''(x)=x-1\), and \(f'''(x)=1\). 2. Solve \(f''(x)=0\), giving \(x=1\). Since \(f'''(1)=1\ne0\), the graph has an inflection point there. 3. Evaluate \(f(1)=\frac{1}{6}-\frac{1}{2}=-\frac{1}{3}\). Thus \(W=\left(1, -\frac{1}{3}\right)\). 4. The tangent slope is \(f'(1)=-\frac{3}{2}\). Using point-slope form, \(y+\frac{1}{3}=-\frac{3}{2}(x-1)\), so \(y=-\frac{3}{2}x+\frac{7}{6}\).

a) \(W=\left(1, -\frac{1}{3}\right)\) b) \(y=-\frac{3}{2}x+\frac{7}{6}\)

- Set the second derivative equal to zero. - Use the third derivative to verify the candidates. - The graph is symmetric about the y-axis. - The points have the same y-coordinate, so compare their x-coordinates.

1. Differentiate: \(f'(x)=4x^3-12x\), \(f''(x)=12x^2-12\), and \(f'''(x)=24x\). 2. Solving \(f''(x)=0\) gives \(x=-1\) and \(x=1\). Since \(f'''(-1)=-24\ne0\) and \(f'''(1)=24\ne0\), both are inflection values. 3. Evaluate the function: \(f(-1)=f(1)=0\). The inflection points are \((-1, 0)\) and \((1, 0)\). 4. The points lie on the same horizontal line, so their distance is \(|1-(-1)|=2\).

The inflection points are \((-1, 0)\) and \((1, 0)\). The distance between them is \(2\) units.

- Use the second derivative to locate the inflection value. - The first derivative gives the tangent slope. - Use point-slope form for the tangent line. - Set the first derivative equal to \(0.9\) for part c.

1. Differentiate: \(g'(x)=-0.3x^2+1.2x\), \(g''(x)=-0.6x+1.2\), and \(g'''(x)=-0.6\). 2. Solve \(g''(x)=0\), giving \(x=2\). Since \(g'''(2)=-0.6\ne0\), the graph has an inflection point there. Also, \(g(2)=2.6\), so \(W=(2, 2.6)\). 3. The tangent slope is \(g'(2)=1.2\). Using point-slope form, \(y-2.6=1.2(x-2)\), so the tangent line is \(y=1.2x+0.2\). 4. Set the derivative equal to the requested slope: \(-0.3x^2+1.2x=0.9\). This simplifies to \(x^2-4x+3=0\), or \((x-1)(x-3)=0\). Thus \(x=1\) or \(x=3\).

a) \(W=(2, 2.6)\) b) \(y=1.2x+0.2\) c) \(x=1\) and \(x=3\)

- Use zeros and sign changes of \(f'\) to classify local extrema. - The function is concave up where its derivative is increasing. - Inflection points of \(f\) correspond to local extrema of \(f'\).

1. The derivative is zero and changes sign at approximately \(x \approx -1.7\), \(x = 0\), and \(x \approx 1.7\). At the first and third values, it changes from negative to positive, giving local minima. At \(x = 0\), it changes from positive to negative, giving a local maximum. 2. The graph of \(f\) is concave up where \(f'\) is increasing. The derivative graph increases on \((-2.5, -1)\) and \((1, 2.5)\). 3. Inflection points of \(f\) occur where \(f'\) has local extrema. These occur at \(x = -1\) and \(x = 1\).

1. Local minima at approximately \(x \approx -1.7\) and \(x \approx 1.7\); local maximum at \(x = 0\) 2. Concave up on \((-2.5, -1)\) and \((1, 2.5)\) 3. \(x = -1\) and \(x = 1\)

- Use the sign of \(f'\) to determine where \(f\) increases or decreases. - At a zero of \(f'\), check how the sign changes to classify a local extremum of \(f\). - Use whether \(f'\) is increasing or decreasing to determine the concavity of \(f\). - Inflection points of \(f\) correspond to local extrema of \(f'\).

1. True. On \((-3, 0)\), the graph of \(f'\) is above the \(x\)-axis, so \(f\) is strictly increasing on \([-3, 0]\). 2. False. At \(x=0\), \(f'\) changes from positive to negative, so \(f\) has a local maximum there. 3. False. The graph of \(f'\) decreases at first on \([1, 3]\), reaches a local minimum between \(x=2\) and \(x=3\), and then increases. Therefore, \(f\) is not concave down throughout the interval. 4. True. Inflection points of \(f\) occur where \(f'\) changes from increasing to decreasing or from decreasing to increasing. The graph of \(f'\) has one local maximum and one local minimum in the displayed interval, so \(f\) has exactly two inflection points there.

1. True 2. False 3. False 4. True

- Find where \(f'\) crosses the \(x\)-axis, and check the sign change at each crossing. - The graph of \(f\) is concave up where the graph of \(f'\) is increasing.

1. The zeros of \(f'\) are \(x=1\) and \(x=3\). At \(x=1\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x=3\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 2. The graph of \(f\) is concave up where \(f'\) is increasing. In the displayed domain, \(f'\) increases for \(-0.5 < x < 2\). Therefore, \(f\) is concave up on \((-0.5, 2)\). At \(x=2\), \(f'\) changes from increasing to decreasing, so \(f\) has an inflection point there.

a) Local minimum at \(x=1\); local maximum at \(x=3\). b) In the displayed domain, \(f\) is concave up on \((-0.5, 2)\) because \(f'\) is increasing there.

- Look for where the graph changes how it bends. - Which derivative determines concavity? - Set the second derivative equal to zero to find interval boundaries. - Test the sign of the second derivative on each interval.

1. From the graph, \(f\) appears concave down between about \(x=-1\) and \(x=1\), and concave up outside that interval. 2. The second derivative is \(f''(x)=3x^2-3=3(x-1)(x+1)\). It is zero at \(x=-1\) and \(x=1\). 3. The second derivative is negative on \((-1, 1)\), so \(f\) is concave down there. It is positive on \((-\infty, -1)\) and \((1, \infty)\), so \(f\) is concave up there.

a) Approximately concave down on \((-1, 1)\) and concave up outside that interval. b) Concave down on \((-1, 1)\); concave up on \((-\infty, -1)\) and \((1, \infty)\).

- Look for where the graph changes how it bends. - Which derivative determines concavity? - Set the second derivative equal to zero. - Check its sign on either side of the candidate value.

1. From the graph, the concavity appears to change near \(x=1\). On the displayed domain, the graph is concave down on \([-3.5, 1)\) and concave up on \((1, 5.5]\). 2. Differentiate: \(f'(x)=x^2-2x-3\), \(f''(x)=2x-2\), and \(f'''(x)=2\). 3. Solving \(f''(x)=0\) gives \(x=1\). Since \(f'''(1)=2\ne0\), this is an inflection value. The second derivative is negative for \(x<1\) and positive for \(x>1\), confirming the graphical estimate.

a) Concave down on \([-3.5, 1)\); concave up on \((1, 5.5]\). b) The concavity changes at \(x=1\); globally, \(f\) is concave down for \(x<1\) and concave up for \(x>1\).

- Find the three values where the graph changes concavity. - Classify the bending separately on each resulting interval.

1. On the displayed domain \([-6, 6]\), the graph changes concavity at \(x=-3\), \(x=0\), and \(x=3\). 2. The graph is concave down on \([-6, -3)\) and \((0, 3)\). 3. The graph is concave up on \((-3, 0)\) and \((3, 6]\).

Concave down on \([-6, -3)\) and \((0, 3)\); concave up on \((-3, 0)\) and \((3, 6]\).

- First estimate the concavity intervals directly from the graph. - Locate where the graph changes how it bends. - Use the second derivative to verify the interval boundaries. - Test the sign of the second derivative in each interval.

1. From the graph, the concavity changes at about \(x=-1\) and \(x=1\). On the displayed domain, the graph is concave up on \([-3.5, -1)\) and \((1, 3.5]\), and concave down on \((-1, 1)\). 2. The second derivative is \(f''(x)=1.5x^2-1.5=1.5(x-1)(x+1)\). 3. It is positive for \(x<-1\) and \(x>1\), and negative for \(-1<x<1\), confirming the graphical result.

a) Concave up on \([-3.5, -1)\) and \((1, 3.5]\); concave down on \((-1, 1)\). b) \(f''(x)=1.5x^2-1.5\), so globally the graph is concave up for \(x<-1\) and \(x>1\), and concave down for \(-1<x<1\).

- Locate where the graph changes concavity. - Compute the second derivative. - Check its sign on either side of its zero.

1. Differentiate: \(f'(x)=0.3x^2-1.2x+0.9\) and \(f''(x)=0.6x-1.2\). 2. Solving \(f''(x)=0\) gives \(x=2\). Also, \(f(2)=1.2\), so the inflection point is \((2, 1.2)\). 3. The second derivative is negative for \(x<2\) and positive for \(x>2\). Thus, on the displayed interval, the graph is concave down on \([-1, 2)\) and concave up on \((2, 5]\).

a) Concave down on \([-1, 2)\); concave up on \((2, 5]\). b) \(f''(x)=0.6x-1.2\), so the inflection value is \(x=2\), with inflection point \((2, 1.2)\).

- A horizontal tangent means the first derivative is zero. - Determine whether the graph is increasing or decreasing at each point. - Concave down means the second derivative is negative; concave up means it is positive. - At an inflection point, concavity changes.

1. At \(A\), the graph has a horizontal tangent and is concave down. Thus, \(f'=0\) and \(f''<0\), so \(A\) is a local maximum. 2. At \(B\), the graph is decreasing and changes from concave down to concave up. Thus, \(f'<0\) and \(f''=0\), so \(B\) is an inflection point. 3. At \(C\), the graph has a horizontal tangent and is concave up. Thus, \(f'=0\) and \(f''>0\), so \(C\) is a local minimum.

\(A\): \(f'=0\), \(f''<0\); local maximum \(B\): \(f'<0\), \(f''=0\); inflection point \(C\): \(f'=0\), \(f''>0\); local minimum

- Use the point's vertical position to determine the sign of \(f\). - Use whether the graph rises or falls to determine the sign of \(f'\). - A horizontal tangent means \(f'=0\). - A change in concavity indicates an inflection point.

1. Point \(A\) is at \(x=\frac{\pi}{2}\). Here, \(f<0\), the graph is decreasing so \(f'<0\), and concavity changes, so \(f''=0\). Thus, \(A\) is an inflection point. 2. Point \(B\) is at \(x=\pi\). Here, \(f<0\), the tangent is horizontal so \(f'=0\), and the graph is concave up so \(f''>0\). Thus, \(B\) is a local minimum. 3. Point \(C\) is at \(x=\frac{3\pi}{2}\). Here, \(f<0\), the graph is increasing so \(f'>0\), and concavity changes, so \(f''=0\). Thus, \(C\) is an inflection point.

\(A\): \(f<0\), \(f'<0\), \(f''=0\); inflection point \(B\): \(f<0\), \(f'=0\), \(f''>0\); local minimum \(C\): \(f<0\), \(f'>0\), \(f''=0\); inflection point

- For a periodic graph, look for crossings of its midline where the bending changes. - Classify the concavity on each interval between consecutive inflection values.

1. From the graph, the concavity changes near \(x=-6.3\), \(x=0\), and \(x=6.3\). 2. The corresponding inflection points are approximately \((-6.3, -1.3)\), \((0, 0)\), and \((6.3, 1.3)\). 3. On the displayed domain, the graph is concave up approximately on \((-6.3, 0)\) and \((6.3, 10]\).

Inflection points: approximately \((-6.3, -1.3)\), \((0, 0)\), and \((6.3, 1.3)\). Concave up approximately on \((-6.3, 0)\) and \((6.3, 10]\).

- Use the sign of \(f'\) to determine where \(f\) increases or decreases. - Check the direction of the sign change in \(f'\) at \(x=1\). - The graph of \(f\) is concave up where \(f'\) is increasing. - Inflection points of \(f\) correspond to local extrema of \(f'\).

1. Statement a) is true. The derivative is negative on \((1, 3)\), so \(f\) is strictly decreasing on \([1, 3]\). 2. Statement b) is false. At \(x=1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 3. Statement c) is false. The graph of \(f'\) has a local maximum between \(x=-1\) and \(x=0\). Thus, \(f'\) changes from increasing to decreasing within \([-1, 1]\), so the concavity of \(f\) changes there. 4. Statement d) is true. Inflection points of \(f\) occur where \(f'\) changes between increasing and decreasing. The graph of \(f'\) has exactly two local extrema in the displayed interval.

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

- Locate where the graph changes concavity. - Look for where the graph is steepest in the part below the x-axis. - On which side are the tangent slopes decreasing? - A change from decreasing slopes to increasing slopes produces what kind of extremum of \(f'\)?

1. The graph changes from concave down to concave up at \(x=2\), so this is the inflection value. 2. On the displayed domain, the graph is concave down on \([-1, 2)\). 3. Before \(x=2\), the slopes decrease; after \(x=2\), the slopes increase. Therefore, \(f'\) has a local minimum at \(x=2\).

1) \(x=2\) 2) \([-1, 2)\) 3) \(f'\) has a local minimum at \(x=2\).

- Identify where the graph bends upward like a cup. - Look for where the graph is steepest downward and steepest upward. - How does the derivative behave when the slopes are becoming larger? - The sign of the second derivative determines concavity.

1. From the graph, the concavity changes at \(x=2\) and \(x=6\). Between these values, the graph bends upward, so \(g\) is concave up on \((2, 6)\). 2. The graph is steepest downward at \(x=2\), so the greatest negative slope occurs there. It is steepest upward at \(x=6\), so the greatest positive slope occurs there. Both values are inflection values because the concavity changes at each one. 3. Where \(g\) is concave down, its slopes decrease, so \(g'\) is decreasing. Where \(g\) is concave up, its slopes increase, so \(g'\) is increasing.

a) \((2, 6)\) b) The greatest negative slope occurs at \(x=2\), and the greatest positive slope occurs at \(x=6\). Both are inflection values. c) On concave-down intervals, \(g'\) decreases. On concave-up intervals, \(g'\) increases.

- Use the first two derivatives to locate and classify the points. - Treat \(k\) as a constant when differentiating. - Eliminate \(k\) separately for each locus. - Use \(k>0\) to determine the attainable x-values.

1. Differentiate: \(f_k'(x)=3x(x-2k)\) and \(f_k''(x)=6x-6k\). 2. Because \(k>0\), the first derivative changes from positive to negative at \(x=0\) and from negative to positive at \(x=2k\). Thus the local maximum is \((0, 0)\), and the local minimum is \((2k, -4k^3)\). 3. Since \(f_k''(x)=6(x-k)\) changes sign at \(x=k\), the inflection point is \((k, -2k^3)\). 4. All local maxima form the single point \((0, 0)\). For the local minima, \(x=2k>0\), so \(k=\frac{x}{2}\) and \(y=-\frac{1}{2}x^3\). For the inflection points, \(x=k>0\), so \(y=-2x^3\).

a) Local maximum: \((0, 0)\); local minimum: \((2k, -4k^3)\); inflection point: \((k, -2k^3)\) b) Maximum locus: \(\{(0, 0)\}\); minimum locus: \(y=-\frac{1}{2}x^3\) for \(x>0\); inflection-point locus: \(y=-2x^3\) for \(x>0\) c) The displayed curves and loci agree with these equations.

- Local minima are the valleys of the graph. - Identify where the graph bends downward like an upside-down bowl. - Count the values where the graph changes from concave up to concave down or vice versa.

1. Differentiate: \(g'(x)=\frac{4}{9}x(x^2-3)\). The critical values are \(x=0\) and \(x=\pm\sqrt{3}\). The graph has local minima at \(x=\pm\sqrt{3}\approx\pm1.73\). 2. The second derivative is \(g''(x)=\frac{4}{3}(x^2-1)\). It is negative when \(-1<x<1\), so the graph is concave down on \((-1, 1)\). 3. The second derivative changes sign at \(x=-1\) and \(x=1\). Therefore, the graph has two inflection points.

a) \(x=-\sqrt{3}\approx-1.73\) and \(x=\sqrt{3}\approx1.73\) b) \((-1, 1)\) c) Two inflection points, at \(x=-1\) and \(x=1\).

- Replace \(x\) with \(-x\) and simplify. - Determine the largest and smallest possible values of the exponential factor. - The steepest points occur at extrema of the first derivative. - Apply the product and chain rules.

1. Since \(g(-x)=4e^{-(-x)^2/4}=4e^{-x^2/4}=g(x)\), the function is even and its graph has y-axis symmetry. 2. Because \(-x^2/4\le0\), the exponential factor satisfies \(0<e^{-x^2/4}\le1\). Therefore, the range is \((0,4]\). 3. Differentiate: \(g'(x)=-2xe^{-x^2/4}\) and \(g''(x)=(x^2-2)e^{-x^2/4}\). Since the exponential factor is positive, \(g''(x)=0\) when \(x=\pm\sqrt{2}\). The second derivative changes sign at both values, so they are inflection-point x-values. The slope reaches its maximum positive and negative values there because \(g'(x)\to0\) as \(x\to\pm\infty\).

a) \(g(-x)=g(x)\), so the graph has y-axis symmetry. b) \((0,4]\). c) \(x=-\sqrt{2}\) and \(x=\sqrt{2}\), approximately \(-1.41\) and \(1.41\).

- Remember that the displayed graph is \(f'\), not \(f\). - Use zeros and sign changes of \(f'\) to classify local extrema of \(f\). - Use the sign of \(f'\) to determine where \(f\) increases or decreases. - Inflection points of \(f\) correspond to local extrema of \(f'\). - The slope of the graph of \(f'\) at a point equals the value of \(f''\) there.

1. Statement (1) is true. At \(x=0\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 2. Statement (2) is true. The derivative is negative on \((0, 3)\), so \(f\) is strictly decreasing on \([0, 3]\). 3. Statement (3) is true. Inflection points of \(f\) occur where \(f'\) changes between increasing and decreasing. The graph of \(f'\) has exactly two local extrema in the displayed interval, so \(f\) has exactly two inflection points there. 4. Statement (4) is true. The graph of \(f'\) is increasing at \(x=2\), so its slope is positive and \(f''(2)>0\). 5. Statement (5) is false. At \(x=3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum.

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

- How does differentiation change the degree of a polynomial? - How many zeros does a nonconstant linear function have? - What must happen to the second derivative at an inflection point? - How many real zeros can a cubic polynomial have, and can all three be sign-changing zeros?

1. Write \(f(x)=ax^3+bx^2+cx+d\), where \(a\ne0\). Then \(f'(x)=3ax^2+2bx+c\) and \(f''(x)=6ax+2b\). Since \(6a\ne0\), \(f''\) is linear. 2. The linear function \(f''(x)=6ax+2b\) has exactly one zero, \(x=-\frac{b}{3a}\). Because its slope is nonzero, \(f''\) changes sign there, so \(f\) has exactly one inflection point. 3. For a fifth-degree polynomial, \(g''\) has degree \(3\), so it can have at most three real zeros and therefore at most three sign changes. This upper bound is attainable. For example, \(g(x)=x^5-\frac{10}{3}x^3\) has \(g''(x)=20x(x-1)(x+1)\), which changes sign at \(x=-1\), \(0\), and \(1\). Thus, the greatest possible number is three.

a) If \(f(x)=ax^3+bx^2+cx+d\), then \(f''(x)=6ax+2b\), which is linear. b) The nonconstant linear function \(f''\) has exactly one zero and changes sign there. c) Three inflection points

- Find a quartic whose second derivative has two simple real zeros. - Relate the parity of \(n\) to the end behavior of the derivative. - Look for a higher-degree polynomial whose second derivative never changes sign.

1. Statement a) is true. For \(f(x)=x^4-6x^2\), \(f''(x)=12x^2-12\), which changes sign at both \(x=-1\) and \(x=1\). 2. Statement b) is true. If \(n\) were even, then \(f'\) would have odd degree and opposite end behavior. Its number of sign-changing zeros would be odd, so \(f\) could not have exactly two local extrema. Therefore, exactly two local extrema require odd \(n\). 3. Statement c) is false. For example, \(f(x)=x^4\) has \(f''(x)=12x^2\geq0\), so its concavity never changes and it has no inflection point.

a) True b) True c) False; for example, \(f(x)=x^4\)

- Think of a basic function with a stationary inflection point. - The graph of \(x^3\) has a horizontal tangent at the origin. - Use translations to move that point to \((-3,2)\). - Verify that the first derivative does not change sign.

1. A horizontal tangent without a local extremum can occur at a stationary inflection point. 2. Shift the basic cubic \(y=x^3\) left \(3\) units and up \(2\) units: \(f(x)=(x+3)^3+2\). 3. Then \(f(-3)=2\) and \(f'(x)=3(x+3)^2\), so \(f'(-3)=0\). 4. Since \(f'(x)\geq0\) on both sides of \(x=-3\), the derivative does not change sign and no local extremum occurs.

One possible function is \(f(x)=(x+3)^3+2\).

- An extremum of \(f'\) is related to a zero and sign change of \(f''\). - A stationary inflection point combines a horizontal tangent with a change in concavity. - A nonzero third derivative alone does not force the second derivative to be zero. - The sign of the third derivative determines how the second derivative crosses zero.

1. Statement a) is true. A sign change in \(f''\) is exactly a change in concavity of \(f\). 2. Statement b) is true. A stationary inflection point has a horizontal tangent, so \(f'(x_0)=0\), and the concavity change means \(f''\) changes sign, making \(f'\) locally extreme. 3. Statement c) is false. The condition \(f'''(x_0)\neq0\) does not require \(f''(x_0)=0\). For example, \(f(x)=x^3+x^2\) has \(f'''(0)=6\), but \(f''(0)=2\), so \(x=0\) is not an inflection point. 4. Statement d) is true. Since \(f''(x_0)=0\) and \(f'''(x_0)<0\), the second derivative changes from positive to negative. Together with \(f'(x_0)=0\), this gives a stationary inflection point with concavity changing from up to down.

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

- Which derivative determines concavity? - Use the chain rule when differentiating the exponential function. - How does the sign of the second derivative determine concavity? - The density function and the factor \(\frac{1}{\sigma^2}\) are always positive.

1. Differentiate using the chain rule: \(\varphi_{\mu,\sigma}'(x)=-\frac{x-\mu}{\sigma^2}\varphi_{\mu,\sigma}(x)\). 2. Differentiate again using the product and chain rules: \(\varphi_{\mu,\sigma}''(x)=\frac{1}{\sigma^2}\left[\left(\frac{x-\mu}{\sigma}\right)^2-1\right]\varphi_{\mu,\sigma}(x)\). 3. Since \(\varphi_{\mu,\sigma}(x)>0\) and \(\sigma^2>0\), the sign of the second derivative is determined by \(\left(\frac{x-\mu}{\sigma}\right)^2-1\). This expression is \(0\) at \(x=\mu-\sigma\) and \(x=\mu+\sigma\). 4. The second derivative is positive when \(x<\mu-\sigma\) or \(x>\mu+\sigma\), and negative when \(\mu-\sigma<x<\mu+\sigma\).

The graph is concave up on \((-\infty, \mu-\sigma)\) and \((\mu+\sigma, \infty)\). It is concave down on \((\mu-\sigma, \mu+\sigma)\).

- Use the point-slope form of a tangent line. - Substitute the inflection point and the derivative at that point. - To find the x-intercept, set the tangent-line expression equal to \(0\). - Simplify common factors before solving for \(x\).

1. The inflection point is \(W\left(\mu+\sigma, \frac{1}{\sigma\sqrt{2\pi e}}\right)\). 2. Since \(\varphi_{\mu,\sigma}'(x)=-\frac{x-\mu}{\sigma^2}\varphi_{\mu,\sigma}(x)\), the slope at \(x=\mu+\sigma\) is \(m=-\frac{1}{\sigma^2\sqrt{2\pi e}}\). 3. Using point-slope form, the tangent line is \(t(x)=-\frac{1}{\sigma^2\sqrt{2\pi e}}(x-\mu-\sigma)+\frac{1}{\sigma\sqrt{2\pi e}}\). 4. Set \(t(x)=0\). Multiplying by \(\sigma^2\sqrt{2\pi e}\) gives \(-(x-\mu-\sigma)+\sigma=0\), so \(x=\mu+2\sigma\).

1. \(t(x)=-\frac{1}{\sigma^2\sqrt{2\pi e}}(x-\mu-\sigma)+\frac{1}{\sigma\sqrt{2\pi e}}\) 2. The x-intercept is \(x=\mu+2\sigma\).

- Use the first- and second-derivative conditions for a local maximum. - Every line in the family passes through the origin. - Compare \(m\) with the secant slope \(\frac{f(x)}{x}\). - Use the concavity of the sine half-wave on the stated interval.

1. Differentiate: \(f'(x)=\frac{\pi}{2}\cos\left(\frac{\pi}{12}x\right)\), so \(f'(6)=0\). 2. The second derivative is \(f''(x)=-\frac{\pi^2}{24}\sin\left(\frac{\pi}{12}x\right)\). Since \(f''(6)=-\frac{\pi^2}{24}<0\), the graph has a local maximum at \(x=6\). 3. The origin is an intersection for every \(m\). On \((0,12]\), intersections satisfy \(\frac{f(x)}{x}=m\). 4. The graph is strictly concave down on \((0,12)\), with \(f(0)=0\). Therefore, the secant slope \(\frac{f(x)}{x}\) decreases strictly from the limiting value \(f'(0)=\frac{\pi}{2}\) to \(0\) at \(x=12\). 5. Hence, when \(0\le m<\frac{\pi}{2}\), there is exactly one positive intersection in addition to the origin. When \(m<0\) or \(m\ge\frac{\pi}{2}\), there is no positive intersection.

a) \(f'(6)=0\) and \(f''(6)=-\frac{\pi^2}{24}<0\), so \(x=6\) is the location of a local maximum. b) For \(0\le m<\frac{\pi}{2}\), there are \(2\) intersection points. For \(m<0\) or \(m\ge\frac{\pi}{2}\), there is \(1\) intersection point.

- A smooth join requires both position and slope to match. - Check the two endpoints separately. - Substitute \(x=1\) to determine the parameter. - Use the second derivative and verify a change in concavity.

1. A smooth connection requires matching function values and slopes. The derivative is \(h_k'(x)=4kx^3-12kx^2+(8k+1)x\). At \(x=0\), \(h_k(0)=0\) and \(h_k'(0)=0\), matching the x-axis. At \(x=2\), \(h_k(2)=2\) and \(h_k'(2)=2\), matching the value and slope of \(y=2x-2\). 2. At \(x=1\), \(h_k(1)=k-4k+4k+0.5=k+0.5\). Setting this equal to \(1\) gives \(k=0.5\). 3. For \(k=1\), \(h_1''(x)=12x^2-24x+9\). Solving \(h_1''(x)=0\) gives \(x=0.5\) and \(x=1.5\). The third derivative is \(h_1'''(x)=24x-24\), which is nonzero at both values, so both are inflection points. Their coordinates are \(W_1=(0.5, 0.6875)\) and \(W_2=(1.5, 1.6875)\).

a) \(h_k(0)=0\), \(h_k'(0)=0\), \(h_k(2)=2\), and \(h_k'(2)=2\), so the values and slopes match at both connections. b) \(k=0.5\) c) The inflection points are \((0.5, 0.6875)\) and \((1.5, 1.6875)\).

- Treat \(k\) as a positive constant. - Remember that an equation \(x^2=c\) may have two solutions, and verify a change in concavity at each candidate. - The derivative value is the tangent slope. - Set the slope expression equal to the required value and solve for \(k\).

1. Differentiate: \(f_k'(x)=2kxe^{1-x^2}\) and \(f_k''(x)=2ke^{1-x^2}(1-2x^2)\). 2. Since the exponential factor is positive, \(f_k'(x)=0\) only at \(x=0\). Because \(f_k''(0)=2ke>0\), the unique local extremum is the local minimum \((0, -ke)\). 3. Inflection-point candidates satisfy \(1-2x^2=0\), so \(x=\pm\frac{1}{\sqrt{2}}\). Since \(1-2x^2\) changes sign at both values, both are x-coordinates of inflection points. 4. At either x-value, \(f_k(x)=-ke^{1/2}=-k\sqrt{e}\). Thus, the inflection points are \(\left(\pm\frac{1}{\sqrt{2}}, -k\sqrt{e}\right)\). 5. At \(x=\frac{1}{\sqrt{2}}\), the slope is \(k\sqrt{2e}\). Setting this equal to \(1\) gives \(k=\frac{1}{\sqrt{2e}}\).

a) Local minimum: \((0, -ke)\) b) \(\left(\pm\frac{1}{\sqrt{2}}, -k\sqrt{e}\right)\) c) \(k=\frac{1}{\sqrt{2e}}\)

- Compare \(g(-x)\) with \(g(x)\). - Symmetric tangent points have opposite slopes and equal y-intercepts. - Set the tangent-line equations equal to find their intersection. - Find where the second derivative changes sign.

1. Since \(g(-x)=4e^{-(-x)^2/4}=g(x)\), the graph is symmetric about the y-axis. 2. The derivative is \(g'(x)=-2xe^{-x^2/4}\). At \(x=2\), \(g(2)=\frac{4}{e}\) and \(g'(2)=-\frac{4}{e}\). Therefore, the tangent line is \(y=-\frac{4}{e}(x-2)+\frac{4}{e}=-\frac{4}{e}x+\frac{12}{e}\). By symmetry, the tangent at \(x=-2\) has slope \(\frac{4}{e}\) and equation \(y=\frac{4}{e}x+\frac{12}{e}\). 3. Set the two line equations equal. This gives \(x=0\), and then \(y=\frac{12}{e}\). Their intersection is \(\left(0,\frac{12}{e}\right)\). 4. As \(x\to\pm\infty\), the exponent approaches \(-\infty\), so \(g(x)\to0\). The second derivative is \(g''(x)=(x^2-2)e^{-x^2/4}\). It changes sign at \(x=\pm\sqrt{2}\). The inflection points are \(\left(-\sqrt{2},\frac{4}{\sqrt{e}}\right)\) and \(\left(\sqrt{2},\frac{4}{\sqrt{e}}\right)\).

a) \(g(-x)=g(x)\), so the graph is symmetric about the y-axis. b) At \(x=2\): \(y=-\frac{4}{e}x+\frac{12}{e}\). At \(x=-2\): \(y=\frac{4}{e}x+\frac{12}{e}\). c) \(\left(0,\frac{12}{e}\right)\). d) \(\lim_{x\to\pm\infty}g(x)=0\). Inflection points: \(\left(-\sqrt{2},\frac{4}{\sqrt{e}}\right)\) and \(\left(\sqrt{2},\frac{4}{\sqrt{e}}\right)\).

- Find the function value and derivative at the origin. - For part b, write the tangent line at a general value \(x=u\), then use the fact that it passes through \(A\). - Remember that an exponential function is never zero. - Use the sign of the second derivative to determine concavity.

1. Since \(g(0)=0\) and \(g'(x)=(1-x)e^{-x}\), the slope at the origin is \(g'(0)=1\). Therefore, the tangent line is \(y=x\). 2. Let the point of tangency be \(B(u, g(u))\). The tangent line at \(x=u\) is \(y=g'(u)(x-u)+g(u)\). Because it passes through \(A(4, 0)\), \(0=(1-u)e^{-u}(4-u)+ue^{-u}\). 3. Since \(e^{-u}\neq0\), simplify to \((1-u)(4-u)+u=0\), or \(u^2-4u+4=0\). Thus \((u-2)^2=0\), so \(u=2\). 4. The point of tangency is \(B(2, 2e^{-2})\), and the slope is \(g'(2)=-e^{-2}\). The tangent line is \(y=-e^{-2}(x-2)+2e^{-2}=-e^{-2}x+4e^{-2}\). 5. The second derivative is \(g''(x)=(x-2)e^{-x}\). Because \(e^{-x}>0\), \(g''(x)<0\) for \(x<2\) and \(g''(x)>0\) for \(x>2\). Therefore, the graph is concave down on \((-\infty, 2)\) and concave up on \((2, \infty)\).

a) \(y=x\) b) \(B(2, 2e^{-2})\); \(y=-e^{-2}x+4e^{-2}\) c) Concave down on \((-\infty, 2)\) and concave up on \((2, \infty)\).

- Convert each width to a radius and each height to units of \(100\,\text{m}\). - Take natural logarithms to solve for the parameters. - The sight line is tangent to the profile at an inflection point. - The top of the tower is the y-intercept of that tangent line.

1. A width of \(200\,\text{m}\) gives radius \(x=1\), and height \(400\,\text{m}\) gives \(h=4\). Similarly, the second condition gives \(x=2\) and \(h=2\). Thus, \(e^{a(b-1)}=4\) and \(e^{a(b-4)}=2\). 2. Taking natural logarithms and subtracting gives \(3a=\ln2\), so \(a=\frac{\ln2}{3}\). Substitution gives \(b=7\). 3. The summit height is \(h(0)=e^{ab}=2^{7/3}\approx5.040\), or about \(504.0\,\text{m}\). 4. The positive inflection point is \(x_W=\sqrt{\frac{1}{2a}}=\sqrt{\frac{3}{2\ln2}}\). Its height is \(h(x_W)=h(0)e^{-1/2}\approx3.057\). 5. Since \(h'(x)=-2axh(x)\), the tangent at \(x_W\) has y-intercept \(-x_Wh'(x_W)+h(x_W)=2h(x_W)\approx6.113\). The tower height is \(6.113-5.040\approx1.073\) units, or about \(107.4\,\text{m}\).

a) \(a=\frac{\ln2}{3}\approx0.231\), \(b=7\) b) About \(107.4\,\text{m}\)

- What derivative condition identifies a local extremum? - How do the first and second derivatives describe slope and concavity? - Check whether every calculated point lies in the modeled interval. - What does it mean geometrically for a horizontal line to pass through a unique maximum?

1. Differentiate: \(f'(x)=(x-0.1x^2)e^{-0.2x}=x(1-0.1x)e^{-0.2x}\). 2. The critical values are \(x=0\) and \(x=10\). The derivative is positive on \((0,10)\) and negative on \((10,15)\), so the maximum occurs at \(x=10\). 3. The maximum height is \(f(10)=50e^{-2}\approx 6.77\,\text{m}\). 4. The second derivative is \(f''(x)=(0.02x^2-0.4x+1)e^{-0.2x}\). Its zeros satisfy \(x^2-20x+50=0\), so \(x=10\pm 5\sqrt{2}\). 5. The left inflection point is \(10-5\sqrt{2}\approx 2.93\), and the right one is \(10+5\sqrt{2}\approx 17.07\). Thus there is no inflection point with \(x>5\) in \([0,15]\); the next change in concavity would occur beyond the modeled barrier. 6. The crest at \(x=10\) is the unique maximum in \([0,15]\). Therefore, every other point to its right has height less than \(50e^{-2}\), so the horizontal ray does not intersect the barrier again in the modeled interval.

a) The maximum height is \(50e^{-2}\,\text{m}\approx 6.77\,\text{m}\) at \(x=10\,\text{m}\). b) There is no inflection point with \(x>5\) in \([0,15]\). The calculated right inflection point is \(x=10+5\sqrt{2}\approx 17.07\), outside the modeled interval; the interval contains the other inflection point at \(x=10-5\sqrt{2}\approx 2.93\). c) The horizontal ray does not intersect the barrier again to the right of the crest within \([0,15]\).

- Use the first derivative to locate the hilltop and the second derivative to study how slope changes. - The steepest descent corresponds to the minimum value of the first derivative. - Model the line of sight as the line segment from the observer to the hilltop. - How does concavity determine whether a curve lies above or below a chord?

1. Differentiate: \(g'(x)=(-0.5x-1)e^{-0.5x}\) and \(g''(x)=0.25x e^{-0.5x}\). 2. Since the exponential factor is positive, \(g'(x)=0\) when \(-0.5x-1=0\), so \(x=-2\). The derivative changes from positive to negative, so the hilltop is \((-2,2e)\approx(-2,5.44)\). 3. The steepest downward slope occurs where \(g'(x)\) is smallest. Since \(g''(x)<0\) for \(x<0\) and \(g''(x)>0\) for \(x>0\), \(g'\) has its minimum at \(x=0\). There, \(g'(0)=-1\). 4. The line from \(P=(-4,0)\) to the hilltop \((-2,2e)\) has slope \(m=\frac{2e}{2}=e\). 5. The profile slope at \(P\) is \(g'(-4)=e^2>e\). Also, \(g''(x)<0\) on \([-4,-2]\), so the graph is concave down and lies above the chord joining \(P\) to the hilltop. The terrain blocks the line of sight, so the observer cannot see the hilltop.

a) The hilltop is \((-2,2e)\approx(-2,5.44)\). b) The steepest downward slope occurs at \(x=0\), where the slope is \(-1\). c) No. The terrain initially rises more steeply than the line of sight, and the concave-down profile lies above that line segment.

- Translate the point at the origin into a condition on the constant term. - A stationary inflection point provides conditions on the function, first derivative, and second derivative. - A local extremum provides a condition on the first derivative. - Solve the resulting linear system for the coefficients.

1. Write \(f(x)=ax^4+bx^3+cx^2+dx+e\). Passing through the origin gives \(e=0\). 2. A stationary inflection point at \((2, 2)\) gives \(f(2)=2\), \(f'(2)=0\), and \(f''(2)=0\). The local-extremum condition gives \(f'(1)=0\). 3. These conditions produce the system \(16a+8b+4c+2d=2\), \(32a+12b+4c+d=0\), \(48a+12b+2c=0\), \(4a+3b+2c+d=0\). Solving gives \(a=-\frac38\), \(b=\frac52\), \(c=-6\), and \(d=6\). 4. Therefore, \(f(x)=-\frac38x^4+\frac52x^3-6x^2+6x\). Here \(f'''(2)=-3\ne0\), so \((2, 2)\) is an inflection point, and \(f''(1)=-\frac32<0\), so \(x=1\) is a local maximum.

\(f(x)=-\frac38x^4+\frac52x^3-6x^2+6x\)

- Compare \(h(-x)\) with \(h(x)\). - Use the first derivative to find and classify extrema. - Factor the numerator of the second derivative. - Verify that the second derivative changes sign at each candidate.

1. Since \(h(-x)=-h(x)\), the graph is symmetric about the origin. Because the denominator's degree is greater than the numerator's degree, \(\lim_{x\to\pm\infty}h(x)=0\), so the horizontal asymptote is \(y=0\). 2. The first derivative is \(h'(x)=\frac{3-x^2}{(x^2+3)^2}\). It is zero at \(x=\pm\sqrt{3}\). 3. The derivative changes from negative to positive at \(x=-\sqrt{3}\), giving a local minimum at \(\left(-\sqrt{3}, -\frac{\sqrt{3}}{6}\right)\). It changes from positive to negative at \(x=\sqrt{3}\), giving a local maximum at \(\left(\sqrt{3}, \frac{\sqrt{3}}{6}\right)\). 4. The second derivative is \(h''(x)=\frac{2x^3-18x}{(x^2+3)^3}=\frac{2x(x-3)(x+3)}{(x^2+3)^3}\). 5. The denominator is always positive, and the numerator has three distinct zeros at \(x=-3\), \(x=0\), and \(x=3\). Its sign changes at each zero, so all three values give inflection points. 6. Evaluating \(h\) gives the points \(\left(-3, -\frac{1}{4}\right)\), \((0, 0)\), and \(\left(3, \frac{1}{4}\right)\).

a) Symmetric about the origin; horizontal asymptote: \(y=0\) b) Local minimum at \(\left(-\sqrt{3}, -\frac{\sqrt{3}}{6}\right)\); local maximum at \(\left(\sqrt{3}, \frac{\sqrt{3}}{6}\right)\) c) Inflection points: \(\left(-3, -\frac{1}{4}\right)\), \((0, 0)\), and \(\left(3, \frac{1}{4}\right)\)

- Examine the end behavior as \(x\to\pm\infty\). - Find the inflection-point x-values in terms of \(k\), then substitute them into \(g_k\). - Use the graph's symmetry to find the tangent intercepts efficiently. - Write the triangle's area in terms of \(k\), then set it equal to \(2\).

1. Since \(k>0\), \(-kx^2\to-\infty\) as \(x\to\pm\infty\). Therefore, \(\lim_{x\to\pm\infty}e^{-kx^2}=0\), so \(y=0\) is a horizontal asymptote. 2. Differentiate: \(g_k'(x)=-2kxe^{-kx^2}\) and \(g_k''(x)=(4k^2x^2-2k)e^{-kx^2}\). 3. Because the exponential factor is positive, \(g_k''(x)=0\) when \(4k^2x^2-2k=0\). Thus \(x=\pm\frac{1}{\sqrt{2k}}\). The factor changes sign at both values, so both are inflection points. 4. At either inflection point, \(g_k\left(\pm\frac{1}{\sqrt{2k}}\right)=e^{-1/2}=\frac{1}{\sqrt{e}}\). 5. Let \(a=\frac{1}{\sqrt{2k}}\). At \(x=a\), the tangent slope is \(-\frac{\sqrt{2k}}{\sqrt{e}}\). This tangent meets the x-axis at \(x=2a=\frac{\sqrt{2}}{\sqrt{k}}\). By symmetry, the other x-intercept is \(-\frac{\sqrt{2}}{\sqrt{k}}\), so the triangle's base is \(\frac{2\sqrt{2}}{\sqrt{k}}\). 6. The two tangent lines meet on the y-axis at height \(\frac{2}{\sqrt{e}}\). Therefore, \(A(k)=\frac{1}{2}\cdot\frac{2\sqrt{2}}{\sqrt{k}}\cdot\frac{2}{\sqrt{e}}=\frac{2\sqrt{2}}{\sqrt{ke}}\). 7. Set the area equal to \(2\): \(\frac{2\sqrt{2}}{\sqrt{ke}}=2\). Then \(ke=2\), so \(k=\frac{2}{e}\).

a) \(\lim_{x\to\pm\infty}g_k(x)=0\), so \(y=0\) is a horizontal asymptote. b) The inflection points have y-coordinate \(\frac{1}{\sqrt{e}}\). c) \(k=\frac{2}{e}\)

- The radicand must be nonnegative. - A product is zero when at least one factor is zero. - Before differentiating at \(x=-2\), determine when that value is in the domain and whether it is an endpoint. - Apply the product and chain rules. - Use the first- and second-derivative conditions for an interior local minimum. - Concavity is determined by the sign of the second derivative.

1. The square root requires \(x+k\geq0\), so the domain is \([-k,\infty)\). 2. The zeros are \(x=-k\) and \(x=\frac{k}{2}\). 3. Differentiate: \(f_k'(x)=\frac{6x+3k}{2\sqrt{x+k}}\) for \(x>-k\). 4. For the point with x-coordinate \(-2\) to be on the graph, \(k\geq2\). If \(k=2\), then \(x=-2\) is the left endpoint, and nearby function values are negative, so the endpoint is a local maximum, not a local minimum. 5. If \(k>2\), then \(x=-2\) is an interior point. A local minimum there requires \(f_k'(-2)=0\), so \(-12+3k=0\) and \(k=4\). 6. The second derivative is \(f_k''(x)=\frac{6x+9k}{4(x+k)^{3/2}}\). For \(k=4\) and \(x=-2\), this is positive, confirming a local minimum. 7. For any \(k>0\) and \(x>-k\), the denominator is positive and \(6x+9k>6(-k)+9k=3k>0\). Therefore, \(f_k''(x)>0\), so the graph is concave up.

a) Domain: \([-k,\infty)\); zeros: \(x=-k\) and \(x=\frac{k}{2}\) b) \(k=4\) c) \(f_k''(x)=\frac{6x+9k}{4(x+k)^{3/2}}>0\) for \(x>-k\).

- Use the chain rule for the first derivative. - Use the quotient rule for the second derivative. - Concavity is determined by the sign of the second derivative. - Express tangent as a quotient of sine and cosine in part c.

1. Differentiate: \(f'(x)=\frac{2x}{x^2+5}\). Using the quotient rule, \(f''(x)=\frac{10-2x^2}{(x^2+5)^2}\). 2. The denominator is always positive. Thus, \(f''(x)>0\) when \(x^2<5\), so the graph is concave up on \((-\sqrt{5}, \sqrt{5})\). It is concave down on \((-\infty, -\sqrt{5})\cup(\sqrt{5}, \infty)\). 3. Since \(h'(x)=-\frac{g'(x)}{g(x)}\), choose \(g(x)=\cos x\). Then \(g'(x)=-\sin x\), and \(h'(x)=-\frac{-\sin x}{\cos x}=\tan x\). Also, \(\cos x>0\) on the stated interval, so the logarithm is defined.

a) \(f'(x)=\frac{2x}{x^2+5}\), \(f''(x)=\frac{10-2x^2}{(x^2+5)^2}\) b) Concave up on \((-\sqrt{5}, \sqrt{5})\); concave down on \((-\infty, -\sqrt{5})\cup(\sqrt{5}, \infty)\) c) One possible choice is \(g(x)=\cos x\).

- Use the sign of the first derivative to establish monotonicity. - Solve \(y=f(x)\) for \(x\) to find the inverse. - Determine the inverse domain from the logarithm restriction. - Concavity follows from the sign of the second derivative.

1. Differentiate: \(f'(x)=-\frac{g'(x)}{g(x)}\). Since \(g(x)>0\) and \(g'(x)>0\) on \(I\), \(f'(x)<0\). Therefore, \(f\) is strictly decreasing on \(I\). 2. Set \(y=-\ln(e^{2x}+1)\) and solve for \(x\): \(e^{-y}=e^{2x}+1\), \(e^{2x}=e^{-y}-1\), \(x=\frac{1}{2}\ln(e^{-y}-1)\). Thus, \(f^{-1}(x)=\frac{1}{2}\ln(e^{-x}-1)\). Its domain requires \(e^{-x}-1>0\), so \(x<0\). 3. Differentiate twice: \(f'(x)=-\frac{2e^{2x}}{e^{2x}+1}\) and \(f''(x)=-\frac{4e^{2x}}{(e^{2x}+1)^2}\). Since this second derivative is negative for every real \(x\), the graph is concave down everywhere.

a) \(f'(x)=-\frac{g'(x)}{g(x)}<0\), so \(f\) is strictly decreasing on \(I\). b) \(f^{-1}(x)=\frac{1}{2}\ln(e^{-x}-1)\), with domain \((-\infty, 0)\) c) \(f''(x)=-\frac{4e^{2x}}{(e^{2x}+1)^2}<0\) for all real \(x\), so \(f\) is concave down everywhere.

- Relate concavity to the monotonic behavior of the first derivative. - Analyze each formula on its own interval. - Check how the first derivative behaves at the point where the formula changes. - A second derivative need not exist at every point of a concave interval.

1. For \(x<0\), \(f''(x)=6x+12\). This is negative for \(x<-2\) and positive for \(-2<x<0\). 2. For \(x>0\), \(f''(x)=4>0\). 3. At \(x=0\), the second derivative does not exist because the one-sided values approach \(12\) and \(4\). However, the first derivative is continuous there: \(f'(x)=3x^2+12x\) for \(x<0\), \(f'(0)=0\), and \(f'(x)=4x\) for \(x>0\). 4. The derivative \(f'\) is increasing throughout \((-2, \infty)\), including across \(x=0\). Therefore, \(f\) is concave up on \((-2, \infty)\) and concave down on \((-\infty, -2)\).

Concave up on \((-2, \infty)\) Concave down on \((-\infty, -2)\)

- Rewrite the absolute value expression piecewise. - Find the second derivative for each branch. - Check whether the intervals can be joined across \(x=0\). - Concave up means the tangent slope is increasing.

1. For \(x\ge0\), \(f(x)=\frac{1}{6}x^3-x^2\), so \(f''(x)=x-2\). Thus, the graph is concave down for \(0<x<2\) and concave up for \(x>2\). 2. For \(x<0\), \(f(x)=-\frac{1}{6}x^3-x^2\), so \(f''(x)=-x-2\). Thus, the graph is concave up for \(x<-2\) and concave down for \(-2<x<0\). 3. At \(x=0\), the one-sided second derivatives both equal \(-2\). Therefore, the concave-down intervals join across \(0\).

Concave up on \((-\infty, -2)\cup(2, \infty)\) Concave down on \((-2, 2)\)

- What is a point where concavity changes called? - What special name is used when an inflection point has a horizontal tangent? - Does the first derivative change sign at the point? - Track how the slope first increases and then decreases while reaching a maximum value of zero.

1. Since \(g'(0)=0\), the graph has a horizontal tangent at \(x=0\). 2. The change from concave up to concave down makes \(x=0\) an inflection value. 3. An inflection point with a horizontal tangent is a stationary inflection point. 4. Because the graph is concave up before \(0\), \(g'\) increases toward \(g'(0)=0\). Because the graph is concave down after \(0\), \(g'\) decreases from \(0\). Thus, \(g'(x)\le0\) on both sides near \(0\), so \(g\) is decreasing on both sides and has no local extremum there.

The point is a stationary inflection point. It is not a local extremum because \(g'\) does not change sign at \(x=0\).

- How does differentiation change the degree of a polynomial? - Compare the end behavior of odd-degree and even-degree polynomials. - What kind of zero of the second derivative produces an inflection point? - Can a linear function change concavity?

1. Each differentiation lowers the degree by one, so if \(f\) has degree \(5\), then \(f''\) has degree \(3\). 2. A cubic polynomial has opposite end behavior as \(x\to-\infty\) and \(x\to\infty\). By continuity, it has at least one real zero. Because the total multiplicity of its real zeros is odd, at least one zero has odd multiplicity and changes sign. 3. In general, if \(f\) has odd degree \(n\ge3\), then \(f''\) has odd degree \(n-2\ge1\). Therefore, \(f''\) has at least one sign-changing zero. 4. A sign change in \(f''\) produces a change in concavity of \(f\), so \(f\) has at least one inflection point. 5. The condition \(n\ge3\) excludes linear polynomials. For \(n=1\), the second derivative is identically zero and the graph has no change in concavity.

a) Degree \(3\); an odd-degree second derivative has opposite end behavior and at least one sign-changing zero. b) For odd \(n\ge3\), \(f''\) has odd degree \(n-2\ge1\), so it changes sign at least once and \(f\) has an inflection point. The case \(n=1\) is a line.

- Find where the second derivative is zero and changes sign. - The third derivative can classify a local extremum of the first derivative. - Evaluate the first derivative to decide whether an inflection point is stationary. - Use exact unit-circle values at \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\).

1. Differentiate three times: \(f'(x)=-\sin x-1\), \(f''(x)=-\cos x\), and \(f'''(x)=\sin x\). 2. Set the second derivative equal to zero: \(-\cos x=0\). On \([0, 2\pi]\), the solutions are \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). The second derivative changes sign at both inputs, so both are inflection points. 3. At \(x=\frac{\pi}{2}\), \(f'''\left(\frac{\pi}{2}\right)=1>0\), so \(f'\) has a local minimum. Also, \(f'\left(\frac{\pi}{2}\right)=-2\ne0\), so this is not a stationary inflection point. Its coordinates are \(\left(\frac{\pi}{2}, -\frac{\pi}{2}\right)\). 4. At \(x=\frac{3\pi}{2}\), \(f'''\left(\frac{3\pi}{2}\right)=-1<0\), so \(f'\) has a local maximum. Also, \(f'\left(\frac{3\pi}{2}\right)=0\), so this is a stationary inflection point. Its coordinates are \(\left(\frac{3\pi}{2}, -\frac{3\pi}{2}\right)\).

\(\left(\frac{\pi}{2}, -\frac{\pi}{2}\right)\): inflection point where the slope has a local minimum; not stationary \(\left(\frac{3\pi}{2}, -\frac{3\pi}{2}\right)\): stationary inflection point where the slope has a local maximum

- What conditions must the first and second derivatives satisfy at a stationary inflection point? - A stationary inflection point is an inflection point with a horizontal tangent. - How can the third derivative confirm a change in concavity? - Substitute each \(x\)-value into the original function.

1. Differentiate: \(f'(x)=\frac{1}{4}x^4-x^3+x^2=\frac{1}{4}x^2(x-2)^2\), \(f''(x)=x^3-3x^2+2x=x(x-1)(x-2)\), and \(f'''(x)=3x^2-6x+2\). 2. A stationary inflection point must satisfy both \(f'(x)=0\) and \(f''(x)=0\). The common zeros are \(x=0\) and \(x=2\). 3. Since \(f'''(0)=2\ne0\) and \(f'''(2)=2\ne0\), the graph changes concavity at both values. 4. Evaluate the function: \(f(0)=0\) and \(f(2)=\frac{32}{20}-\frac{16}{4}+\frac{8}{3}=\frac{4}{15}\).

\((0, 0)\) and \(\left(2, \frac{4}{15}\right)\)

- Look for a function whose second derivative changes sign while its first derivative stays positive. - Origin symmetry means the polynomial is odd. - Differentiate an odd polynomial twice and analyze the result near \(0\). - An inflection point requires a concavity change.

1. Statement a) is false. Let \(f(x)=\frac{1}{5}x^5-\frac{2}{3}x^3+5x\). Then \(f''(x)=4x(x^2-1)\), which changes sign at \(x=-1\), \(0\), and \(1\). However, \(f'(x)=x^4-2x^2+5=(x^2-1)^2+4>0\), so the function has no local extrema. 2. Statement b) is true. An origin-symmetric polynomial is odd, so its second derivative is also odd. Because the polynomial has degree at least \(3\), the second derivative is not identically zero and its lowest nonzero power is odd. Therefore, it changes sign at \(x=0\), giving an inflection point.

a) False; \(f(x)=\frac{1}{5}x^5-\frac{2}{3}x^3+5x\) is a counterexample. b) True

- Use the second derivative to locate an inflection point and check its sign on both sides. - Analyze the discriminant of the first derivative to determine when it has fewer than two distinct real zeros. - Interpret the sign of the first derivative in the cost context.

1. Differentiate: \(K'(x) = 0.3x^2 - 3x + c\), \(K''(x) = 0.6x - 3\), and \(K'''(x) = 0.6\). 2. Solve \(K''(x) = 0\): \(0.6x - 3 = 0\), so \(x = 5\). Because \(K''(x) < 0\) for \(x < 5\) and \(K''(x) > 0\) for \(x > 5\), the graph changes from concave down to concave up at \(x = 5\), independently of \(c\). 3. Interior local extrema occur where \(K'(x)\) changes sign. If the discriminant of \(K'(x) = 0.3x^2 - 3x + c\) is negative, then \(K'(x) > 0\) for all \(x\). If the discriminant is zero, the derivative has one double zero and does not change sign. If the discriminant is positive, the two distinct real zeros have sum \(10\), so at least one is positive; each simple zero changes the sign of the derivative and gives an interior local extremum. Thus, there are no interior local extrema exactly when the discriminant is nonpositive. 4. Compute \(D = (-3)^2 - 4(0.3)c = 9 - 1.2c\). Solve \(9 - 1.2c \le 0\): \(c \ge 7.5\). 5. For \(c \ge 7.5\), \(K'(x) \ge 0\) for all \(x\), with at most one isolated zero. Thus, total cost never decreases as production increases; there is no interior production level at which total cost changes from increasing to decreasing or vice versa.

a) The inflection point occurs at \(x = 5\), where the graph changes from concave down to concave up. b) \(c \ge 7.5\) c) Total cost does not decrease as production rises; it has no interior local maximum or minimum.

- A stationary inflection point on the x-axis corresponds to a zero of multiplicity \(3\). - Use a factored quartic form with one additional zero. - Apply the point and slope conditions at \(x=0\). - Expand the final expression only after determining the parameters.

1. A stationary inflection point on the x-axis at \(x=2\) is a zero of multiplicity \(3\). Write \(f(x)=a(x-2)^3(x-k)\). 2. The condition \(f(0)=8\) gives \(8ak=8\), so \(ak=1\). 3. Differentiating gives \(f'(x)=a(x-2)^2(4x-3k-2)\). Thus, \(f'(0)=-12ak-8a=-20\). Using \(ak=1\) gives \(a=1\), and then \(k=1\). 4. Therefore, \(f(x)=(x-2)^3(x-1)=x^4-7x^3+18x^2-20x+8\). Also, \(f'''(2)=6\ne0\), confirming the stationary inflection point.

\(f(x)=x^4-7x^3+18x^2-20x+8\)

- The exponential factor is never zero. - A horizontal tangent requires the first derivative to equal zero. - Use the discriminant to count the real roots of each quadratic factor. - An inflection point requires a sign change in the second derivative.

1. Since \(e^x>0\), zeros come from \(x^2-k=0\). If \(k<0\), there are no real zeros. If \(k=0\), \(x=0\) is a double zero. If \(k>0\), the zeros \(x=\pm\sqrt{k}\) are both simple. 2. By the product rule, \(g_k'(x)=(x^2+2x-k)e^x\). The condition \(g_k'(1)=0\) gives \((3-k)e=0\), so \(k=3\). 3. Critical points satisfy \(x^2+2x-k=0\). This quadratic has two distinct real roots when its discriminant \(4+4k\) is positive, so exactly two local extrema occur when \(k>-1\). 4. The second derivative is \(g_k''(x)=(x^2+4x+2-k)e^x\). The quadratic factor has no sign-changing zeros when its discriminant \(8+4k\) is nonpositive. For \(k<-2\), there are no real zeros; for \(k=-2\), there is a double zero without a sign change. Therefore, there are no inflection points when \(k\le-2\).

a) \(k<0\): no real zeros; \(k=0\): \(x=0\), multiplicity \(2\); \(k>0\): \(x=\pm\sqrt{k}\), each with multiplicity \(1\) b) \(k=3\) c) \(k>-1\) d) \(k\le-2\)

- Check whether the denominator can be zero and compare \(f(-x)\) with \(f(x)\). - Use the numerator for x-intercepts and leading coefficients for the horizontal asymptote. - Evaluate the second derivative at the intercept x-values and verify sign changes. - Use the first and second derivatives to classify the critical point.

1. Since \(x^2+3>0\) for every real \(x\), the domain is \(\mathbb{R}\). Since \(f(-x)=f(x)\), the graph is symmetric about the y-axis. 2. The numerator is zero at \(x=\pm 1\), so the x-intercepts are \((-1, 0)\) and \((1, 0)\). Equal numerator and denominator degrees give the horizontal asymptote \(y=1\). 3. The first derivative is \(f'(x)=\frac{8x}{(x^2+3)^2}\), and the second derivative is \(f''(x)=\frac{24(1-x^2)}{(x^2+3)^3}\). 4. The second derivative is zero at \(x=\pm 1\). Its denominator is positive, and the numerator changes sign at both values, so \((-1, 0)\) and \((1, 0)\) are inflection points. 5. The first derivative is zero only at \(x=0\). Since \(f''(0)>0\), the graph has a local minimum at \(\left(0, -\frac{1}{3}\right)\).

1) Domain: \(\mathbb{R}\); symmetric about the y-axis 2) x-intercepts: \((-1, 0)\) and \((1, 0)\); horizontal asymptote: \(y=1\) 3) Both x-intercepts are inflection points. 4) Local minimum: \(\left(0, -\frac{1}{3}\right)\)

- Set the second derivative equal to zero and isolate one exponential expression. - Determine when the right side of the resulting exponential equation is positive. - Use the third derivative to verify a change in concavity. - Reuse the inflection-point equation to simplify the y-coordinate.

1. The second derivative is \(f_a''(x)=ae^x+a^2e^{ax}=a(e^x+ae^{ax})\). 2. Since \(a\ne0\), an inflection point must satisfy \(e^x=-ae^{ax}\), or \(e^{(1-a)x}=-a\). This equation has a real solution exactly when \(-a>0\), so exactly when \(a<0\). 3. The solution is \(x=\frac{\ln(-a)}{1-a}\). The third derivative at this point is \(a(1-a)e^x\ne0\), so the second derivative changes sign. 4. From \(e^{(1-a)x}=-a\), we have \(e^x=(-a)^{\frac{1}{1-a}}\). Also, \(e^{ax}=-\frac{1}{a}e^x\). Therefore, the y-coordinate is \(\left(a-\frac{1}{a}\right)e^x=\frac{a^2-1}{a}(-a)^{\frac{1}{1-a}}\).

The inflection point exists exactly when \(a<0\), and its coordinates are \(\left(\frac{\ln(-a)}{1-a}, \frac{a^2-1}{a}(-a)^{\frac{1}{1-a}}\right)\).

- Use the product rule and chain rule. - Set the first and second derivatives equal to zero. - Horizontal distance is the difference between x-coordinates. - An exponential equals \(1\) only when its exponent is zero.

1. The first two derivatives are \(g_a'(x)=(1-a^2-ax)e^{-ax}\) and \(g_a''(x)=(a^2x+a^3-2a)e^{-ax}\). 2. Since the exponential factor is positive, \(g_a'(x)=0\) gives the local-maximum x-coordinate \(x=\frac{1}{a}-a\). The derivative changes from positive to negative there. 3. Similarly, \(g_a''(x)=0\) gives the inflection-point x-coordinate \(x=\frac{2}{a}-a\). 4. The horizontal distance is \(\left(\frac{2}{a}-a\right)-\left(\frac{1}{a}-a\right)=\frac{1}{a}\). 5. The slope at the inflection point is \(-e^{a^2-2}\). Setting this equal to \(-1\) gives \(e^{a^2-2}=1\), so \(a^2=2\). Since \(a>0\), \(a=\sqrt{2}\).

a) Local-maximum x-coordinate: \(\frac{1}{a}-a\); inflection-point x-coordinate: \(\frac{2}{a}-a\) b) The horizontal distance is \(\frac{1}{a}\). c) \(a=\sqrt{2}\)

- Find the first three derivatives. - Solve the first- and second-derivative equations separately on \(x>0\). - Track how the sign of \(k\) affects whether each equation has a real solution. - Use logarithm properties to simplify the y-coordinate.

1. The first three derivatives are \(g_k'(x)=\frac{k}{x}+2x\), \(g_k''(x)=-\frac{k}{x^2}+2\), and \(g_k'''(x)=\frac{2k}{x^3}\). 2. A critical point satisfies \(2x^2=-k\). Since \(x>0\), this has exactly one solution when \(k<0\): \(x=\sqrt{-\frac{k}{2}}\). At this point, \(g_k''(x)=4>0\), so it is a local minimum. 3. An inflection point satisfies \(2x^2=k\). This has exactly one positive solution when \(k>0\): \(x=\sqrt{\frac{k}{2}}\). Since \(g_k'''(x)\ne0\) there, the concavity changes. 4. The y-coordinate is \(k\ln\left(\sqrt{\frac{k}{2}}\right)+\frac{k}{2}=\frac{k}{2}\left(\ln\left(\frac{k}{2}\right)+1\right)\).

1. For \(k<0\), there is exactly one local minimum. For \(k>0\), there is exactly one inflection point. 2. For \(k>0\), the inflection point is \(\left(\sqrt{\frac{k}{2}}, \frac{k}{2}\left(\ln\left(\frac{k}{2}\right)+1\right)\right)\).

- The exponential factor never equals zero. - Compare \(f(-x)\) with \(f(x)\) and \(-f(x)\). - Factor out the exponential term after applying the product rule. - Use the absolute minimum and end behavior to determine the range.

1. Since the exponential factor is never zero, the x-intercepts satisfy \(x^2-4=0\), giving \((-2,0)\) and \((2,0)\). Also, \(f(0)=-4\), so the y-intercept is \((0,-4)\). 2. Neither \(f(-x)=f(x)\) nor \(f(-x)=-f(x)\), so the graph has neither y-axis symmetry nor origin symmetry. 3. As \(x\to\infty\), \(f(x)\to\infty\). As \(x\to-\infty\), exponential decay dominates polynomial growth, so \(f(x)\to0\). 4. The first derivative is \(f'(x)=\left(\frac{1}{2}x^2+2x-2\right)e^{x/2}\). Its zeros are \(x=-2\pm2\sqrt{2}\). The second derivative is \(f''(x)=\left(\frac{1}{4}x^2+2x+1\right)e^{x/2}\). It is negative at \(x=-2-2\sqrt{2}\) and positive at \(x=-2+2\sqrt{2}\). Thus the local maximum is \(\left(-2-2\sqrt{2},8(1+\sqrt{2})e^{-1-\sqrt{2}}\right)\approx(-4.83,1.73)\), and the local minimum is \(\left(-2+2\sqrt{2},8(1-\sqrt{2})e^{-1+\sqrt{2}}\right)\approx(0.83,-5.01)\). 5. Solving \(f''(x)=0\) gives \(x=-4\pm2\sqrt{3}\). The concavity changes at both values. The inflection points are \(\left(-4-2\sqrt{3},8(3+2\sqrt{3})e^{-2-\sqrt{3}}\right)\approx(-7.46,1.24)\) and \(\left(-4+2\sqrt{3},8(3-2\sqrt{3})e^{-2+\sqrt{3}}\right)\approx(-0.54,-2.84)\). 6. The local minimum is the absolute minimum, and the function is unbounded above. Therefore, the range is \(\left[8(1-\sqrt{2})e^{-1+\sqrt{2}},\infty\right)\).

1) x-intercepts: \((-2,0)\), \((2,0)\); y-intercept: \((0,-4)\). 2) Neither y-axis symmetry nor origin symmetry. 3) \(\lim_{x\to\infty}f(x)=\infty\) and \(\lim_{x\to-\infty}f(x)=0\). 4) Local maximum: \(\left(-2-2\sqrt{2},8(1+\sqrt{2})e^{-1-\sqrt{2}}\right)\); local minimum: \(\left(-2+2\sqrt{2},8(1-\sqrt{2})e^{-1+\sqrt{2}}\right)\). 5) Inflection points: \(\left(-4-2\sqrt{3},8(3+2\sqrt{3})e^{-2-\sqrt{3}}\right)\) and \(\left(-4+2\sqrt{3},8(3-2\sqrt{3})e^{-2+\sqrt{3}}\right)\). 6) \(\left[8(1-\sqrt{2})e^{-1+\sqrt{2}},\infty\right)\).

- Find the inflection points in terms of \(a\). - Use the horizontal segment between them as the base. - The height is the absolute value of their y-coordinate. - Express both sides of the final equation as powers with the same exponent.

1. Differentiate: \(f_a'(x)=4x^3-2ax\) and \(f_a''(x)=12x^2-2a\). The inflection-point equation gives \(x=\pm\sqrt{\frac{a}{6}}\). Since the third derivative is nonzero at both values, they are inflection points. 2. Their y-coordinate is \(\frac{a^2}{36}-\frac{a^2}{6}=-\frac{5a^2}{36}\). Thus the horizontal base of the triangle has length \(2\sqrt{\frac{a}{6}}\), and its height is \(\frac{5a^2}{36}\). 3. The area is \(A=\frac{1}{2}\cdot2\sqrt{\frac{a}{6}}\cdot\frac{5a^2}{36}=\frac{5a^{5/2}}{36\sqrt{6}}\). 4. Setting \(A=5\) gives \(a^{5/2}=36\sqrt{6}=6^{5/2}\). Because \(a>0\), \(a=6\).

\(a=6\)

- Find the inflection points from the second derivative. - Use symmetry to identify the triangle’s base. - The common y-coordinate gives the height. - Solve the resulting equation in \(k^{3/2}\).

1. Differentiate twice: \(f_k'(x)=-\frac{2k^2x}{(x^2+k)^2}\) and \(f_k''(x)=\frac{2k^2(3x^2-k)}{(x^2+k)^3}\). 2. The inflection-point equation gives \(x=\pm\sqrt{\frac{k}{3}}\). The second derivative changes sign at both values, so they are inflection points. 3. Their y-coordinate is \(\frac{k^2}{k/3+k}=\frac{3k}{4}\). The triangle therefore has base \(2\sqrt{\frac{k}{3}}\) and height \(\frac{3k}{4}\). 4. Its area is \(A=\frac{1}{2}\cdot2\sqrt{\frac{k}{3}}\cdot\frac{3k}{4}=\frac{\sqrt{3}}{4}k^{3/2}\). 5. Setting this equal to \(2\sqrt{3}\) gives \(k^{3/2}=8\), so \(k=4\).

\(k=4\)

- Use the product rule twice. - Solve the second-derivative equation for \(k\). - Substitute that expression into the original function. - Translate \(k>-2\) into a restriction on \(x\). - Verify the candidates using the third derivative.

1. By the product rule, \(f_k'(x)=(x^2+2x-k)e^x\) and \(f_k''(x)=(x^2+4x+2-k)e^x\). 2. Since \(e^x>0\), an inflection point must satisfy \(k=x^2+4x+2\). 3. Substituting this relation into the original function gives \(y=\left(x^2-(x^2+4x+2)\right)e^x=(-4x-2)e^x\). 4. The parameter restriction requires \(x^2+4x+2>-2\), or \((x+2)^2>0\). Therefore, \(x\neq-2\). 5. At a candidate point, the third derivative is \(2(x+2)e^x\), which is nonzero precisely because \(x\neq-2\). Thus every allowed point is an inflection point.

\(y=(-4x-2)e^x\) for \(x\neq-2\)

- Apply the quotient rule, then use \(\sin^2x+\cos^2x=1\). - A square is always nonnegative. - Differentiate the first-derivative formula to study concavity. - An inflection point requires a change in concavity, not only a zero of the second derivative.

1. Apply the quotient rule: \(f'(x)=\frac{(\cos x)(\cos x)-(\sin x)(-\sin x)}{\cos^2x}\) \(=\frac{\cos^2x+\sin^2x}{\cos^2x}\) \(=1+\frac{\sin^2x}{\cos^2x}=1+\tan^2x\). 2. Since \(\tan^2x\ge 0\), \(f'(x)=1+\tan^2x\ge 1>0\). Therefore, \(f\) is strictly increasing on the entire interval. 3. Differentiate the first derivative: \(f''(x)=2\tan x(1+\tan^2x)\). Because \(1+\tan^2x>0\), \(f''(x)=0\) only when \(\tan x=0\), which gives \(x=0\). The factor \(\tan x\) changes sign at \(0\), so the concavity changes there. Since \(f(0)=0\), the inflection point is \((0,0)\).

1. \(f'(x)=1+\tan^2x\) 2. \(f'(x)>0\) throughout the interval, so \(f\) is strictly increasing. 3. Inflection point: \((0,0)\)

- Translate the stationary-inflection conditions into equations involving the function and its derivatives. - Factor the resulting polynomial and its first derivative. - Compare the formulas for the nonzero zero and critical point. - Consider the degree and possible multiplicities of zeros of the first derivative.

1. A stationary inflection point at the origin requires \(p(0)=0\), \(p'(0)=0\), and \(p''(0)=0\). Thus \(e=d=c=0\), so \(p(x)=ax^4+bx^3\). Also, \(p'''(0)=6b\neq0\), so \(b\neq0\). 2. Factor the polynomial: \(p(x)=x^3(ax+b)\). In addition to the triple zero at the origin, the only other zero has x-coordinate \(-\frac{b}{a}\). 3. Differentiate: \(p'(x)=x^2(4ax+3b)\). The origin is the stationary inflection point, and the only other critical point is \(x=-\frac{3b}{4a}\). Because the linear factor changes sign there, this point is a local extremum. 4. The requested ratio is \(\frac{-b/a}{-3b/(4a)}=\frac{4}{3}\). 5. A stationary inflection point is a zero of multiplicity at least two of \(p'\). Since the derivative of a fourth-degree polynomial has degree three, it cannot have two distinct zeros that each have multiplicity at least two. Therefore, a fourth-degree polynomial cannot have two distinct stationary inflection points.

a) \(c=d=e=0\), so \(p(x)=ax^4+bx^3\), where \(a\neq0\) and \(b\neq0\) b) Additional zero: \(x=-\frac{b}{a}\); local extremum: \(x=-\frac{3b}{4a}\) c) The ratio is \(\frac{4}{3}\). d) Two stationary inflection points would require the cubic derivative to have two distinct double zeros, which is impossible.

- Factor the second derivative and determine when the quadratic equation has real solutions. - Use the third derivative to verify each candidate. - Eliminate \(k\) from the nonzero inflection-point condition. - Compare the sizes of \(x^3\) and \(x^5\) near \(x=0\).

1. The second and third derivatives are \(g_k''(x)=2x(10kx^2-3)\) and \(g_k'''(x)=6(10kx^2-1)\). 2. The second derivative is zero at \(x=0\) for every \(k\). If \(k\leq0\), the equation \(10kx^2=3\) has no real solution, so the origin is the only inflection point. If \(k>0\), there are two additional candidates \(x=\pm\sqrt{\frac{3}{10k}}\). 3. At the origin, \(g_k'''(0)=-6\neq0\). At either nonzero candidate, the third derivative equals \(12\neq0\). Thus there is one inflection point when \(k\leq0\) and three when \(k>0\). 4. For a nonzero inflection point, \(k=\frac{3}{10x^2}\). Substitution gives \(y=\frac{3}{10x^2}x^5-x^3=-\frac{7}{10}x^3\). 5. Derivative argument: \(g_k''(0)=0\) and \(g_k'''(0)=-6\neq0\) for every \(k\), so the origin is always an inflection point. 6. Qualitative argument: near the origin, the fifth-degree term \(kx^5\) is much smaller in magnitude than the cubic term \(-x^3\). Therefore, the graph locally has the same concavity change as \(y=-x^3\).

a) One inflection point for \(k\leq0\); three inflection points for \(k>0\) b) \(y=-\frac{7}{10}x^3\) for the nonzero inflection points c) The origin is always an inflection point because \(g_k''(0)=0\) and \(g_k'''(0)=-6\neq0\); near the origin, \(-x^3\) dominates \(kx^5\).

- What do the tangent line’s value and slope tell you about \(f(1)\) and \(f'(1)\)? - Translate each graph condition into an equation involving \(f\), \(f'\), or \(f''\). - How many coefficients does a general cubic polynomial have? - Verify the tangent-point condition after finding the coefficients.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Then \(f'(x)=3ax^2+2bx+c\) and \(f''(x)=6ax+2b\). 2. Translate the conditions into equations: \(f(0)=2\), so \(d=2\). \(f(2)=0\), so \(8a+4b+2c=-2\). \(f''(1)=0\), so \(6a+2b=0\). The tangent slope is \(-3\), so \(f'(1)=-3\), giving \(3a+2b+c=-3\). Also, \(f(1)=t(1)=1\), so \(a+b+c=-1\). 3. From \(6a+2b=0\), \(b=-3a\). Substituting into the remaining equations gives \(-2a+c=-1\) and \(-3a+c=-3\). Subtracting yields \(a=2\). Then \(b=-6\), \(c=3\), and \(d=2\). 4. Therefore, \(f(x)=2x^3-6x^2+3x+2\). Verification gives \(f(2)=0\), \(f(1)=1\), \(f'(1)=-3\), and \(f''(1)=0\). Since \(f'''(x)=12\ne0\), the graph changes concavity at \(x=1\).

\(f(x)=2x^3-6x^2+3x+2\)

- Read the time-coordinate of the highest point on the graph. - At an interior maximum, the first derivative is zero. - The most rapid decrease occurs where the derivative reaches its minimum. - Use the second derivative to locate a possible inflection point.

1. The graph has a maximum at about \(t=4\) hours. Since \(f_k'(t)=4e^{-kt}(1-kt)\), the maximum satisfies \(1-4k=0\), so \(k=0.25\). 2. In general, \(f_k'(t)=0\) when \(1-kt=0\), so \(t_{\max}=\frac{1}{k}\). Also, \(f_k''(t)=4ke^{-kt}(kt-2)\), and \(f_k''\left(\frac{1}{k}\right)=-4ke^{-1}<0\), confirming a maximum. 3. The concentration decreases most rapidly where its slope is minimal, which occurs at an inflection point of \(f_k\). Setting \(f_k''(t)=0\) gives \(kt-2=0\), so \(t=\frac{2}{k}\). Since \(\frac{2}{k}=2\left(\frac{1}{k}\right)\), this time is \(2t_{\max}\).

a) Maximum at about \(4\) hours; \(k=0.25\) b) \(t_{\max}=\frac{1}{k}\) c) The concentration decreases most rapidly at \(t=\frac{2}{k}=2t_{\max}\)

- Differentiate the exponential terms using the chain rule. - Solve exponential equations by isolating an exponential expression and taking logarithms. - For the improper integral, evaluate a limit as the upper bound approaches infinity. - Multiply the concentration and time units to obtain the integral's units.

1. a) \(c'(t)=25(-0.05e^{-0.05t}+0.25e^{-0.25t})\). Setting \(c'(t)=0\) gives \(e^{0.2t}=5\), so \(t=5\ln5\approx8.05\,\text{min}\). 2. b) \(c''(t)=25(0.0025e^{-0.05t}-0.0625e^{-0.25t})\). Setting \(c''(t)=0\) gives \(e^{0.2t}=25\), so \(t=5\ln25=10\ln5\approx16.09\,\text{min}\). This is the inflection point on the decreasing portion of the graph, where \(c'(t)\) is least. 3. c) \(B=25\left[-20e^{-0.05t}+4e^{-0.25t}\right]_0^{\infty}=25(16)=400\,\frac{\text{mg}\cdot\text{min}}{\text{L}}\).

a) \(t=5\ln5\,\text{min}\approx8.05\,\text{min}\) b) \(t=10\ln5\,\text{min}\approx16.09\,\text{min}\) c) \(B=400\,\frac{\text{mg}\cdot\text{min}}{\text{L}}\)

- Study the sign of \(\sin(x)-e^{-x}\) at key points. - Use concavity to limit the number of zeros in \([0, \pi]\). - Identify intervals where \(\sin(x)\leq0\). - Repeat the argument on each positive half-wave of sine.

1. Let \(h(x)=\sin(x)-e^{-x}\). Then \(h(0)=-1<0\), \(h\!\left(\frac{\pi}{2}\right)=1-e^{-\pi/2}>0\), and \(h(\pi)=-e^{-\pi}<0\). The Intermediate Value Theorem gives at least two zeros in \([0, \pi]\). 2. On \((0, \pi)\), \(h''(x)=-\sin(x)-e^{-x}<0\), so \(h\) is strictly concave. A strictly concave function cannot cross the \(x\)-axis more than twice when it is negative at both endpoints and positive at an interior point. Thus, there are exactly two solutions. 3. On \([\pi, 2\pi]\), \(\sin(x)\leq0\) while \(e^{-x}>0\), so there is no solution. 4. For every nonnegative integer \(k\), on \([2k\pi, (2k+1)\pi]\), the sine function is \(0\) at both endpoints and \(1\) at the midpoint, while \(e^{-x}\) is positive and less than \(1\) at the midpoint. Therefore, there are two solutions in each such interval. The equation has infinitely many solutions.

a) Exactly two solutions in \([0, \pi]\) b) No solutions in \([\pi, 2\pi]\) c) Infinitely many solutions for \(x\geq0\)