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

- Positive and negative second derivatives correspond to opposite concavity directions. - An inflection point requires a change in concavity at a point on the graph.

1. Since \(f''(x)>0\) for \(x<3\), the graph is concave up on \((-\infty,3)\). 2. Since \(f''(x)<0\) for \(x>3\), the graph is concave down on \((3,\infty)\). 3. The concavity changes at the continuous point \(x=3\), so \(x=3\) is the x-coordinate of an inflection point.

Concave up on \((-\infty,3)\); concave down on \((3,\infty)\); inflection-point x-coordinate \(3\)

- Compare how steeply the curve rises on the left and right. - Concavity describes how tangent slopes change, not just whether the curve rises. - A curve can rise and still become steeper.

1. The derivative is \(f'(x)=e^x\), which increases as \(x\) increases. 2. Thus the tangent slopes become larger from left to right. 3. Increasing slopes mean the graph is concave up.

The tangent slopes get larger, and the graph is concave up.

- This graph is the derivative, not the function itself. - Look at the direction in which the derivative curve moves on the positive side. - Does a decreasing derivative mean the original function's slopes grow or shrink?

1. The plotted derivative \(-x^2\) decreases as \(x\) runs from \(0\) to \(2\). 2. A decreasing first derivative means the slopes of \(f\) are getting smaller. 3. Therefore \(f\) is concave down on \((0,2)\).

\(f'\) is decreasing; \(f\) is concave down on \((0,2)\).

- Decide whether the second derivative is positive or negative. - Connect its sign to the direction in which tangent slopes change.

1. The second derivative is positive throughout \((-3,2)\). 2. A positive second derivative means the tangent slopes of \(f\) are increasing. 3. Hence \(f\) is concave up on \((-3,2)\).

Concave up on \((-3,2)\), because \(f''(x)>0\) there.

- Translate each second-derivative sign into a concavity direction. - What must change at an inflection point? - Check that the function does not jump at the point.

1. A positive second derivative makes the graph concave up on the left. 2. A negative second derivative makes it concave down on the right. 3. Because the function is continuous and the concavity changes, \(x=2\) is an inflection point.

The graph changes from concave up to concave down, so \(x=2\) is an inflection point.

- A zero second derivative marks a possible inflection point, not a guaranteed one. - Determine the second derivative's sign for a negative and a positive x-value. - Does the sign change?

1. The second derivative is positive on both sides of \(0\) and zero only at \(0\). 2. The graph is concave up on both sides, so its concavity does not change. 3. Therefore \(0\) is not an inflection point.

No. The second derivative has the same positive sign on both sides of \(0\), so there is no concavity change.

- Translate the sign of \(f''\) on either side into concavity. - Does an inflection point require the second derivative to exist at the point itself? - Check that the curve joins continuously at \(0\).

1. The graph is concave down for \(x<0\) and concave up for \(x>0\). 2. The function is continuous at \(0\), and the concavity changes there. 3. Therefore \(0\) is an inflection point even though \(f''(0)\) does not exist.

Yes. \(x=0\) is an inflection point because concavity changes from down to up across that point.

- A horizontal tangent does not necessarily mean a peak or valley. - Does the curve keep rising through \(x=1\)? - Check how its bending changes across the horizontal tangent.

1. The curve rises on both sides of \(x=1\), so it has no local maximum or minimum there. 2. Its concavity changes from down to up at \(x=1\), since \(f''(x)=6(x-1)\). 3. The tangent is horizontal, so \((1,2)\) is a stationary inflection point.

A stationary inflection point at \((1,2)\), not a local extremum.

- Could the derivative become positive between the listed inputs under the stated assumption? - Try sketching tangent lines whose slopes change as the table indicates. - Consider the curve's direction and how its steepness changes as separate questions.

1. The derivative stays negative, so \(f\) is decreasing. 2. The derivative increases, so the slopes are becoming less negative; \(f\) is concave up.

\(f\) is decreasing and concave up.

- Check both sides of the zero of \(f''\). - A zero of the second derivative is only a possible inflection point. - Decide whether the bending direction actually reverses.

1. The second derivative is zero at \(0\), but positive on both sides. 2. The graph stays concave up as it passes through \(0\). 3. Since concavity does not change, \(0\) is not an inflection-point x-coordinate.

No. \(f''(x)\) is positive on both sides of \(0\), so concavity does not change.

- Check whether the derivative is positive or negative. - Check whether the derivative itself rises or falls. - A function can rise while its slope becomes smaller.

1. Since \(5-x>0\) for \(0<x<4\), \(f\) is increasing. 2. Since \(f'\) is decreasing and \(f''=-1<0\), the graph is concave down.

Increasing and concave down throughout \((0,4)\).

- Connect growth rate with the slope of a height graph. - Focus on how that slope changes over time. - A positive rate and an increasing rate answer different questions.

1. The growth rate is the first derivative of height with respect to time. 2. Because the first derivative increases, the height graph is concave up over the stated period.

Concave up. The plant's height increases at an increasing rate.

- 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.

- 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: \(f(1)=-1+3-3+2=1\).

\((1, 1)\)

- 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 \(3\)

- 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, Sora's claim is correct: a line has \(f''(x)=0\) everywhere, so it is neither concave up nor concave down.

- Which derivative represents the rate at which views increase? - What does Fermat's theorem say about an interior maximum of a differentiable function? - Concavity can be read from whether the first derivative increases or decreases.

1. The growth rate is \(V'\). It increases before \(t_w\) and decreases afterward, so it has a local maximum there. The graph of \(V\) changes from concave up to concave down, making \((t_w,V(t_w))\) an inflection point. 2. Because \(V'\) is differentiable and has an interior local maximum, Fermat's theorem gives \(V''(t_w)=0\). 3. An increasing first derivative means \(V\) is concave up before \(t_w\), while a decreasing first derivative means it is concave down afterward. Where the second derivative exists, this gives \(V''(t)\ge0\) before and \(V''(t)\le0\) after. Strict increase of \(V'\) does not require strict positivity of \(V''\) at every point.

1. An inflection point. 2. \(V''(t_w)=0\). 3. Concave up before \(t_w\); concave down after \(t_w\).

- 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 \(\left(3, \frac{7}{4}\right)\).

- 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) \(W=(2, 1)\)

- Which derivative determines concavity? - What equation gives possible inflection-point x-coordinates? - 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 \([-3, 5]\), the graph changes concavity at \(x=1\). It is concave down on \([-3, 1)\) and concave up on \((1, 5]\). 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 \([-3, 1)\); concave up on \((1, 5]\). b) \(W\approx(1, 0.9)\) c) \(f''(-1)<0\) and \(f''(3)>0\)

- 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 \([-3, 3]\), the graph changes concavity near \(x=-1\) and \(x=1\). 2. It is concave up on \([-3, -1)\) and \((1, 3]\), 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 \([-3, -1)\) and \((1, 3]\); 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) \(W\approx(2, 2.6)\)

- 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 x-values give inflection points. 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 x-value gives an inflection point. Evaluating gives \(g(2)=\frac{10}{3}\).

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

- 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 x-values give inflection points. 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.

- Differentiate the exponential function twice. - Use the sign of the second derivative to determine concavity. - Use the sign of the second derivative to determine whether the first derivative increases or decreases.

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

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

- The first derivative gives the ramp's slope. - A horizontal tangent has slope \(0\). - Use higher derivatives to distinguish an extremum from a stationary inflection point. - Check the sign of the first derivative on both sides.

1. The derivatives are \(f'(x)=(x-2)^2\), \(f''(x)=2x-4\), and \(f'''(x)=2\). 2. A horizontal tangent requires \(f'(x)=0\), so \(x=2\). 3. At \(x=2\), \(f'(2)=0\), \(f''(2)=0\), and \(f'''(2)=2\ne0\). Therefore, the point is a stationary inflection point, not a local extremum. 4. Since \(f'(x)=(x-2)^2\ge0\) on the interval, the ramp rises before and after \(x=2\); it is only momentarily horizontal there.

a) \(x=2\,\text{m}\) b) It is a stationary inflection point. The ramp remains increasing on both sides and is momentarily horizontal at \(x=2\).

- 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 \((1, \infty)\).

- 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 \((-2, 2)\).

- 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 \((3, 6]\).

- Determine the sign of \(\sin(x)\) on each side of \(0\). - The sign of \(f''(x)\) is the opposite of the sign of \(\sin(x)\).

1. The second derivative is \(f''(x)=-2\sin(x)\). 2. For \(-\pi<x<0\), \(\sin(x)<0\) and \(f''(x)>0\), so the graph is concave up. 3. For \(0<x<\pi\), \(\sin(x)>0\) and \(f''(x)<0\), so the graph is concave down. At \(x=-\pi\), \(x=0\), and \(x=\pi\), the second derivative is zero; the strict sign statements do not apply at these points. 4. The concavity changes at \(x=0\), giving concave-up interval \([-\pi,0)\) and concave-down interval \((0,\pi]\) on the stated domain.

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 \(x=2\). 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-point x-coordinate 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-point x-coordinates: approximately \(x=-1.4\) and \(x=1.4\). Concave up on \((-1.4, 1.4)\); concave down on \([-5, -1.4)\) and \((1.4, 5]\).

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

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 \((6.3, 10]\).

- Locate the two valleys and estimate their x-coordinates from the grid. - Identify where the graph changes from bending upward to bending downward or vice versa. - Each concavity change gives an inflection point.

1. The two valleys occur at approximately \(x=-1.7\) and \(x=1.7\), so these are the local-minimum x-values. 2. The graph changes concavity near \(x=-1\) and \(x=1\). Between those values, it bends downward, so it is concave down on approximately \((-1, 1)\). 3. The concavity changes at approximately \(x=-1\) and \(x=1\). Therefore, the graph has two inflection points.

a) \(x\approx-1.7\) and \(x\approx1.7\) b) Approximately \((-1, 1)\) c) Two inflection points, at approximately \(x=-1\) and \(x=1\).

- What is \(F''\) when \(F'=f\)? - Inflection points occur where the second derivative changes sign. - Look for local extrema of \(f\).

1. Since \(F'=f\), the second derivative of \(F\) is \(F''=f'\). 2. The concavity of \(F\) changes wherever \(f'\) changes sign, at a local maximum or minimum of \(f\). 3. From the graph, \(f\) has a local minimum near \(x=-2.3\) and a local maximum near \(x=2.3\). Therefore, \(F\) changes concavity at those two x-values.

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

- Do not use whether \(f'\) is above or below the x-axis. - Concavity depends on whether the derivative itself is increasing or decreasing. - Identify the interval over which the derivative graph rises.

1. The graph of \(f\) is concave up where \(f'\) is increasing. 2. The derivative graph is increasing from its local minimum at \(x=0\) to its local maximum at \(x=2\). 3. Therefore, \(f\) is concave up on \((0,2)\).

\((0,2)\).

- For concavity, focus on whether the derivative graph rises or falls rather than whether it is above or below the x-axis. - Separate the domain where the slope behavior of the derivative changes. - An inflection point requires a change in the resulting concavity.

1. The graph of \(f'\) increases on \((-6,-2)\), decreases on \((-2,1)\), increases on \((1,4)\), and decreases on \((4,6)\). 2. Therefore, \(f\) is concave up on \((-6,-2)\) and \((1,4)\), and concave down on \((-2,1)\) and \((4,6)\). 3. The concavity changes at \(x=-2\), \(x=1\), and \(x=4\), so these are the x-coordinates of the inflection points.

Concave up: \((-6,-2)\) and \((1,4)\). Concave down: \((-2,1)\) and \((4,6)\). Inflection-point x-coordinates: \(x=-2,1,4\).

- An inflection point needs a change of concavity, not just a zero second derivative. - Find where the second derivative changes sign. - Remember that the requested answer is a point, so find its function value.

1. Differentiate twice: \(f''(x)=6x-6\). 2. The second derivative is negative when \(x<1\) and positive when \(x>1\), so \(x=1\) is an inflection location. 3. Evaluate \(f(1)=1-3+2=0\), giving the point \((1,0)\).

\((1,0)\); concavity changes from down to up at \(x=1\).

- Look for a familiar trigonometric function whose sign changes at zero. - Differentiate twice to check which sign the second derivative has. - Test the left and right intervals separately.

1. Choose \(f(x)=\sin x\). 2. Its second derivative is \(f''(x)=-\sin x\). 3. Since \(\sin x<0\) on \((-\pi,0)\), \(f''>0\) there; since \(\sin x>0\) on \((0,\pi)\), \(f''<0\) there. 4. Thus the function has exactly the requested concavity on the two intervals.

One example is \(f(x)=\sin x\), since \(f''(x)=-\sin x\) is positive to the left of \(0\) and negative to the right on the given intervals.

- Acceleration is the second derivative of position. - Determine where the acceleration is positive and negative. - Connect acceleration signs to both concavity and changes in velocity.

1. The acceleration is positive for \(0<t<2\), so \(s\) is concave up and its velocity \(s'\) increases. 2. The acceleration is negative for \(2<t<4\), so \(s\) is concave down and its velocity decreases. 3. The acceleration changes sign at \(t=2\), giving an inflection point in the position graph then.

Concave up on \((0,2)\), concave down on \((2,4)\), with an inflection at \(t=2\) hours. Velocity increases before \(2\) hours and decreases afterward.

- Concavity of \(f\) is linked to whether \(f'\) rises or falls. - Look for local maxima and minima of the derivative curve. - A change in direction of \(f'\) corresponds to a change in the sign of \(f''\).

1. The first derivative increases until \(x=\frac{\pi}{2}\), then decreases, and increases again after \(x=\frac{3\pi}{2}\). 2. Thus \(f''(x)=\cos x\) changes sign at both \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). 3. These are the inflection locations of \(f\). The derivative graph does not supply the corresponding values of \(f\).

\(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\).

- Mark the zeros of the second derivative first. - Test one input in each interval. - An inflection point requires a change in concavity.

1. The second derivative is zero at \(x=-2\) and \(x=1\). 2. Its sign is positive on \(( -\infty,-2)\), negative on \((-2,1)\), and positive on \((1,\infty)\). 3. Therefore, \(f\) is concave up outside the two zeros and concave down between them. The sign changes at both zeros, so both are inflection-point x-coordinates.

Concave up: \(( -\infty,-2)\) and \((1,\infty)\). Concave down: \((-2,1)\). Inflection-point x-coordinates: \(-2\) and \(1\).

- Use the sign of the second derivative on the two sides of zero. - An inflection point need not have a horizontal tangent. - The given function and derivative values supply its position and slope.

1. The second derivative is positive for \(x<0\) and negative for \(x>0\). 2. Hence the graph is concave up on \(( -\infty,0)\) and down on \((0,\infty)\). 3. Concavity changes at \(x=0\), so \((0,2)\) is an inflection point. The tangent slope there is \(f'(0)=5\), which is not horizontal.

a) Concave up on \(( -\infty,0)\); concave down on \((0,\infty)\). b) Inflection point \((0,2)\), with tangent slope \(5\).

- How would the original curve bend if its tangent slopes were becoming smaller? - Compare how the plotted slopes change on either side of the lowest point. - What evidence would show an actual concavity change rather than just a stationary point?

1. On \((-2,0)\), the graph of \(f'\) falls, so \(f\) is concave down. 2. On \((0,2)\), the graph of \(f'\) rises, so \(f\) is concave up. 3. The direction of change of \(f'\) switches at \(x=0\), so \(f\) has an inflection point there.

Concave down on \((-2,0)\); concave up on \((0,2)\); inflection-point x-coordinate \(0\).

- 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-point x-coordinate 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-point x-coordinates: \(f(-1)=\frac{1}{12}-\frac{1}{2}-2=-\frac{29}{12}\) 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 \(x>\sqrt{-\frac{1}{t}}\).

- 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-point x-coordinate.

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-point x-coordinate 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-point x-coordinates? - 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 x-values correspond to inflection points. 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.

- Decide which transformations change x-coordinates and which change only y-coordinates. - A vertical reflection or vertical stretch preserves the input at which concavity changes. - For a horizontal scaling, determine which new input produces the original input value.

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 it 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 x-coordinate of the inflection point, so it remains \(x=3\). 3. The function \(r(x)=f(3x)\) compresses the graph horizontally by a factor of \(\frac13\). The original input \(3\) occurs when \(3x=3\), so the new x-coordinate 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 \(\frac13\).

- Use the second derivative to locate the x-coordinate of the inflection point. - Treat the parameter \(t\) as a constant while differentiating with respect to \(x\). - Substitute the x-coordinate into the original function to obtain the full point.

1. Differentiate: \(g_t'(x)=\frac12x^2-2tx+10\), \(g_t''(x)=x-2t\), and \(g_t'''(x)=1\). 2. Solve \(g_t''(x)=0\): \(x-2t=0\), so the x-coordinate of the inflection point is \(x=2t\). Since \(g_t'''(x)=1\ne0\), the graph changes concavity there. 3. Evaluate the function: \(g_t(2t)=\frac16(2t)^3-t(2t)^2+10(2t)=-\frac83t^3+20t\).

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

- 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 \(x=5\).

\(a=-2\)

- First locate the x-coordinate where the graph changes concavity. - A point on the x-axis has y-coordinate \(0\). - Substitute the inflection-point x-coordinate into the original function and impose that condition.

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 x-coordinate of the inflection point is \(x=1\). Since \(g_k'''(x)=6\ne0\), the graph changes concavity there. 3. For the inflection point to lie on the x-axis, its y-coordinate must be \(0\). Thus require \(g_k(1)=0\): \(1-3+k=0\), giving \(k=2\).

\(k=2\)

- For part a), look for a quartic whose second derivative never changes sign. - In part b), consider a stationary point where the derivative has the same sign on both sides. - For part c), consider the degree and slope of the second derivative of a cubic.

1. Statement a) is false. For \(f(x)=x^4+1\), we have \(f''(x)=12x^2\), which does not change sign, so there is no inflection point. 2. Statement b) is false. For \(f(x)=x^3\), \(f'(x)=3x^2\) is zero at \(x=0\) but stays nonnegative on both sides. The function has no local extremum. 3. Statement c) is true. Every cubic has the form \(f(x)=ax^3+bx^2+cx+d\), with \(a\ne0\). Its second derivative is \(f''(x)=6ax+2b\), a nonconstant linear function that changes sign exactly once.

a) False; \(f(x)=x^4+1\) has no inflection point. b) False; \(f(x)=x^3\) has no local extremum. c) True; a cubic has exactly one inflection point.

- 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 x-coordinate of the inflection point in terms of the parameter. - Use the derivative that controls concavity to locate the change. - A point with y-coordinate \(6\) must satisfy the original function equation. - Substitute the inflection-point x-coordinate and solve for the parameter.

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 point has x-coordinate \(x=a\). 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 \(a=-1\).

\(a=-1\)

- First locate the x-coordinates where concavity changes. - Horizontal distance depends only on the x-coordinates of the two points. - Express that distance in terms of the parameter. - Use the given distance to solve for \(k\).

1. Differentiate: \(g_k'(x)=\frac13x^3-kx+5\) and \(g_k''(x)=x^2-k\). 2. Solve \(g_k''(x)=0\): the inflection points have x-coordinates \(x=\pm\sqrt{k}\). Since \(k>0\), both are real and \(g_k'''(x)=2x\ne0\) at each one. 3. The horizontal distance between the two inflection points is the difference of their x-coordinates: \(\sqrt{k}-(-\sqrt{k})=2\sqrt{k}\). 4. Set \(2\sqrt{k}=6\). Then \(\sqrt{k}=3\), so \(k=9\).

\(k=9\)

- Locate the inflection point from the second derivative, then find its y-coordinate from the original function. - Check whether that point satisfies the given 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 \(x=1\); because \(f'''(1)\ne0\), the graph changes concavity there. 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 \(P(0,5)\) moves \(2\) units right and \(4\) units down. Therefore, \(h(x)=f(x-2)-4\). 4. Expanding gives \(h(x)=(x-2)^3-3(x-2)^2-9(x-2)+1=x^3-9x^2+15x-1\).

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

- Locate the x-coordinate where the second derivative changes sign. - Substitute that x-coordinate into the original function to obtain the inflection point. - Which derivative gives the tangent slope? - Use the point and slope to write the tangent line.

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 \(y=x+\frac{4}{3}\)

- 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-point x-coordinate?

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.

- 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 \(x=1\)

- 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.

- Differentiate the logarithm first, then simplify before differentiating again. - Concavity is determined by the sign of the second derivative. - The denominator of \(f''\) is always positive, so focus on its numerator.

1. Differentiate: \(f'(x)=\frac{2x}{x^2+5}\). 2. Using the quotient rule, \(f''(x)=\frac{10-2x^2}{(x^2+5)^2}\). 3. The denominator is always positive. Thus, \(f''(x)>0\) when \(x^2<5\), so the graph is concave up on \((-\sqrt{5}, \sqrt{5})\). 4. The graph is concave down on \((-\infty, -\sqrt{5})\cup(\sqrt{5}, \infty)\).

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 \((-\infty, -\sqrt{5})\cup(\sqrt{5}, \infty)\)

- Think about how differentiation changes the degree of a polynomial. - How many real zeros can the first and second derivatives have? - An upper bound alone is not enough: find a polynomial whose second derivative has four sign-changing zeros.

1. The first derivative of a sixth-degree polynomial has degree \(5\), so it has at most \(5\) real zeros. Each local extremum must occur at one of those critical numbers, so there can be at most \(5\) local extrema. 2. The second derivative has degree \(4\), so it has at most \(4\) real zeros. Each inflection point requires a concavity change, so there can be at most \(4\) inflection points. 3. This bound is attained by \(f(x)=\frac{1}{30}x^6-\frac{5}{12}x^4+2x^2\). Differentiating twice gives \(f''(x)=x^4-5x^2+4=(x^2-1)(x^2-4)\). The second derivative changes sign at all four simple zeros \(x=-2,-1,1,2\). Thus the greatest possible number of inflection points is \(4\).

a) At most \(5\) local extrema. b) At most \(4\) inflection points, and \(4\) is attainable.

- 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-point x-coordinate, which does not depend on \(k\). - Require the tangent slope to be zero at that x-coordinate. - 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-point x-coordinate is \(x=-1\). Since \(g_k'''(-1)=2\ne0\), this is an inflection-point x-coordinate 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 \(\left(-1, -\frac{1}{3}\right)\)

- 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 number 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-point x-coordinate.

a) \(f'(x)=4x^3-12x^2+12x\); \(f''(x)=12x^2-24x+12\) b) A local minimum at \((0,0)\) 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-point x-coordinate. 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-point x-coordinate because \(g''\) does not change sign there.

- 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 \((-2, 2)\)

- 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 \((0, 0)\).

- Is the function increasing or decreasing? - If a decline slows, does the slope become more negative or less negative? - Relate the sign of the second derivative to concavity. - What type of point separates intervals of different concavity?

1. A decline means the stock price is decreasing, so \(f'(t)<0\). 2. A slowing decline means the negative slope is becoming less negative. Therefore, the slope is increasing and \(f''(t)>0\). 3. Since \(f''(t)>0\), the graph is concave up during the slowing decline. A transition from \(f''<0\) to \(f''>0\) is an inflection point and corresponds to the most negative slope.

1. \(f'(t)<0\) 2. \(f''(t)>0\) 3. The graph is concave up. The transition point is an inflection point.

- 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 gives an inflection-point x-coordinate. 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-point x-coordinate \(x=6\) 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-point x-coordinates. 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.

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

- 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.

- 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.

- 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. \(\left(-\ln(4a), \frac{3}{16a}\right)\)

- 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 \(k=1\).

- 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: \((0,0)\)

- 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) \(y=-3x+11\)

- 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 numbers \(x=0\) 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-point x-coordinates? - Use the sign of the second derivative on each interval. - Substitute the inflection-point x-coordinates 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 \((1, \infty)\).

- 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 \((-1, 1)\).

- An inflection point is where the graph changes concavity. - How do the slopes change as you move through a concave-up or concave-down part of the curve? - Can a turning point for concavity be the endpoint of the largest interval with that concavity? - Keep the displayed domain in mind when naming maximal intervals.

1. The second derivative is \(f''(x)=12(x-1)(x-3)\). It changes sign at \(x=1\) and \(x=3\), so those are the inflection-point x-coordinates. 2. At \(x=1\), concavity changes from up to down, so the slope \(f'\) changes from increasing to decreasing and has a local maximum there. In fact, \(f'(1)=8\). 3. At \(x=3\), concavity changes from down to up, so \(f'\) has a local minimum there, with \(f'(3)=-8\). 4. On the displayed domain, the largest concave-up intervals are \([-0.5,1]\) and \([3,4.5]\). The largest concave-down interval is \([1,3]\). The inflection points can be included as endpoints of these intervals even though the concavity changes across them.

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 \([1,3]\).

- Which derivatives locate and verify an inflection point? - What quantity represents the instantaneous rate of change of volume? - Where does the slope reach a maximum when concavity changes from up to down? - Interpret the signs of the first and second derivatives separately.

1. Differentiate: \(V'(t)=-1.5t^2+12t\), \(V''(t)=-3t+12\), and \(V'''(t)=-3\). 2. Solve \(V''(t)=0\): \(-3t+12=0\), so \(t=4\). Since \(V'''(4)=-3\ne0\), the graph has an inflection point there. Also, \(V(4)=84\), so the point is \((4, 84)\). On the graph, this is where the slope stops increasing and begins decreasing. 3. The rate \(V'(t)\) reaches its maximum at \(t=4\). Thus the maximum rate is \(V'(4)=24\,\text{m}^3/\text{h}\). 4. For \(0\le t<4\), \(V''(t)>0\), so the graph is concave up and the rate of change is increasing. For \(4<t\le10\), \(V''(t)<0\), so the graph is concave down and the rate of change is decreasing. Since \(V'(8)=0\), the volume increases through \(t=8\) and decreases after \(t=8\).

a) \((4, 84)\); the volume’s rate of increase is greatest there. b) \(24\,\text{m}^3/\text{h}\) c) Concave up for \(0\le t<4\) and concave down for \(4<t\le10\). The rate of change increases before \(t=4\) and decreases afterward; the volume begins decreasing after \(t=8\).

- Use the sign of the first derivative to determine where the function increases or decreases. - Use the sign of the second derivative to determine the graph's concavity. - Does an isolated zero of a derivative prevent a turning point from being included as the endpoint of a maximal interval?

1. The derivative is \(f'(x)=\frac{3}{4}(x-2)(x+2)\). It is positive for \(x<-2\) and \(x>2\), and negative for \(-2<x<2\). Hence \(f\) 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 \(x>0\). Hence the graph is concave down on \(( -\infty,0]\) and concave up on \([0,\infty)\). The inflection point at \(x=0\) may be included as an endpoint of each maximal concavity interval.

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

- Differentiate each function three times. - Zeros of the second derivative are candidates for inflection-point x-coordinates. - 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-point x-coordinates. 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-point x-coordinate. Also, \(g(2)=4\).

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

- Use the second derivative to locate the inflection-point x-coordinate. - 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) \(y=-\frac{3}{2}x+\frac{7}{6}\)

- Use the second derivative to locate the inflection-point x-coordinate. - 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 \(x=3\)

- Identify each horizontal tangent and decide whether the graph changes direction there. - The sign of \(f'\) is determined by whether the function is increasing or decreasing. - A stationary inflection point has a horizontal tangent and a change in concavity without a change from increasing to decreasing or vice versa.

1. From the graph, the local minimum is approximately \((-5,-2.0)\), the local maximum is approximately \((-1,2.0)\), and the stationary inflection point is approximately \((3,0.8)\). 2. Near \(x=-5\), the graph changes from decreasing to increasing, so \(f'\) changes from negative to positive. 3. Near \(x=-1\), the graph changes from increasing to decreasing, so \(f'\) changes from positive to negative. 4. Near \(x=3\), the graph is decreasing on both sides of the horizontal tangent, so \(f'\) stays negative on both sides. The concavity changes there, making the point a stationary inflection point rather than a local extremum.

a) Local minimum \(\approx(-5,-2.0)\); local maximum \(\approx(-1,2.0)\); stationary inflection point \(\approx(3,0.8)\) b) At \(x=-5\), \(f'\) changes from negative to positive. At \(x=-1\), \(f'\) changes from positive to negative. At \(x=3\), \(f'\) stays negative on both sides.

- Look for where the graph changes how it bends. - Which derivative determines concavity? - Set the second derivative equal to zero, then check its sign on either side.

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 \(f'''(1)=2\ne0\), the graph has an inflection point at \(x=1\). 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 \(x>1\).

- 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 x-coordinate of the inflection point is \(2\), and the inflection point is \((2,1.2)\).

- Where is the graph steepest? - What does an increasing slope mean for population growth? - What resource limits might affect a wildlife population? - Relate concavity to whether the growth rate is increasing or decreasing.

1. The population grows fastest at the graph’s inflection point, approximately \(t=7.5\) years. At that time, the population is about \(50\) eagles. 2. Before \(t=7.5\), the graph is concave up: the growth rate is increasing, so the population grows faster each year. After \(t=7.5\), the graph is concave down: the population still grows, but its growth rate decreases. 3. The leveling off suggests a carrying capacity. Limited food, nesting sites, territory, or other habitat resources can prevent unlimited population growth.

a) About \(7.5\) years, with approximately \(50\) eagles. b) The growth rate increases before \(t=7.5\) and decreases afterward. The inflection point marks the maximum growth rate. c) Resource limits and habitat carrying capacity can cause the population to level off.

- Locate where the graph changes concavity. - Track how the tangent slopes change as x moves through the graph. - 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 concavity changes from down to up at \(x=2\), so the inflection point has x-coordinate \(2\). 2. Since \(f''(x)=0.6(x-2)<0\) for \(x<2\), the graph is concave down on the largest interval \([-1,2]\) in the displayed domain. The inflection point can be included as the endpoint. 3. The tangent slopes decrease before \(x=2\) and increase after \(x=2\). 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 are x-coordinates of inflection points 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 x-coordinates of inflection points. c) On concave-down intervals, \(g'\) decreases. On concave-up intervals, \(g'\) increases.

- Look for where the graph has its greatest positive slope. - Think about which resources in a petri dish are limited.

1. The area grows most rapidly where the displayed curve is steepest. From the graph, the steepest part of the S-shaped curve is centered at about \(t=10\) hours, so the maximum growth rate occurs then. 2. The horizontal line \(h\) is at about \(500\,\text{mm}^2\). The covered area could approach this limit because of limited space, limited nutrients, or other environmental constraints in the petri dish.

a) After \(10\) hours. b) Possible reasons include limited space, limited nutrients, or other limiting environmental conditions.

- What happens to the formula when you replace \(x\) by \(-x\)? - Differentiate twice and determine where the second derivative changes sign. - How does the sign of the second derivative tell you where the first derivative increases or decreases? - Use the graph's symmetry to compare the steepest upward and downward slopes.

1. Since \(g(-x)=4e^{-(-x)^2/4}=g(x)\), the function is even and the graph is symmetric about the y-axis. 2. Differentiate: \(g'(x)=-2xe^{-x^2/4}\) and \(g''(x)=(x^2-2)e^{-x^2/4}\). 3. The exponential factor is positive, so \(g''=0\) at \(x=\pm\sqrt{2}\). It is negative on \((-\sqrt{2},\sqrt{2})\) and positive on \((-\infty,-\sqrt{2})\) and \((\sqrt{2},\infty)\). Both zeros are inflection-point x-values. 4. The derivative \(g'\) increases where \(g''>0\) and decreases where \(g''<0\). Therefore, \(g'\) has its greatest positive value at \(x=-\sqrt{2}\) and its smallest negative value at \(x=\sqrt{2}\).

a) \(g(-x)=g(x)\), so the graph is symmetric about the y-axis. b) Inflection-point inputs: \(x=\pm\sqrt{2}\). Concave down on \((-\sqrt{2},\sqrt{2})\); concave up on \((-\infty,-\sqrt{2})\) and \((\sqrt{2},\infty)\). c) Greatest positive slope at \(x=-\sqrt{2}\); smallest negative slope at \(x=\sqrt{2}\).

- Differentiate the exponential terms using the chain rule. - Solve each exponential equation by isolating an exponential expression and taking logarithms. - The concentration decreases most rapidly where \(c'(t)\) is smallest; what derivative locates an extremum of \(c'\)?

1. For part 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. For part b, \(c''(t)=25(0.0025e^{-0.05t}-0.0625e^{-0.25t})\). Setting \(c''(t)=0\) gives \(e^{0.2t}=25\), so \(t=5\ln25=10\ln5\approx16.09\,\text{min}\). 3. The concavity changes at this time, and on the decreasing portion of the graph \(c'(t)\) reaches its minimum there. Therefore, the concentration is decreasing most rapidly at \(t=10\ln5\,\text{min}\).

a) \(t=5\ln5\,\text{min}\approx8.05\,\text{min}\) b) \(t=10\ln5\,\text{min}\approx16.09\,\text{min}\); this is the inflection-point time, where \(c'\) is minimal.

- A fraction with a fixed positive numerator is largest when its denominator is smallest. - Substitute \(t=12\) into the model. - At an inflection point, the rate of change reaches a local extremum.

1. The numerator is constant, so \(k(t)\) is greatest when the denominator is smallest. The denominator \((t-10)^2+12\) is minimized at \(t=10\). 2. The maximum concentration is \(k(10)=\frac{80}{12}=\frac{20}{3}\approx6.67\,\text{mg/L}\). 3. At the inflection point, \(k(12)=\frac{80}{(12-10)^2+12}=\frac{80}{16}=5\,\text{mg/L}\). 4. On the decreasing side of the curve, the inflection point is where \(k'(t)\) is most negative. Thus, at \(t=12\), the medication concentration is decreasing at its fastest rate.

a) Maximum at \(t=10\) hours: \(\frac{20}{3}\approx6.67\,\text{mg/L}\) b) \(k(12)=5\,\text{mg/L}\). At that time, the concentration is decreasing at its fastest rate.

- Locate where the second derivative changes sign, including points between rows of the table. - Positive and negative second-derivative values correspond to different concavity. - A zero of the second derivative matters here only when the sign actually changes.

1. On \([-3,-1]\), the linear second derivative changes from \(2\) to \(-2\), so it is zero at \(x=-2\). It is positive before \(-2\) and negative after. 2. The second derivative stays negative from \(x=-1\) through \(x=2\). On \([2,5]\), it rises from \(-2\) to \(4\) and is zero at \(x=3\). 3. Therefore, \(f''>0\) on \((-3,-2)\) and \((3,5)\), and \(f''<0\) on \((-2,3)\). The sign changes at \(x=-2\) and \(x=3\), so both are inflection-point x-coordinates.

Concave up on \((-3,-2)\) and \((3,5)\). Concave down on \((-2,3)\). Inflection-point x-coordinates: \(x=-2\) and \(x=3\).

- Read concavity from whether the derivative graph is increasing or decreasing. - Focus on what changes as you pass through the corner. - Distinguish a condition that guarantees or detects an inflection point from the definition of an inflection point.

1. The graph of \(f'\) decreases on \((-4,0)\), so \(f\) is concave down there. It increases on \((0,4)\), so \(f\) is concave up there. 2. The concavity changes from down to up at \(x=0\). Therefore, \(f\) has an inflection point at \(x=0\). The second derivative need not exist at the inflection point; the defining feature is the change in concavity.

a) Concave down on \((-4,0)\); concave up on \((0,4)\). b) Yes. \(x=0\) is the x-coordinate of an inflection point because the concavity changes there, even though \(f''(0)\) does not exist.

- 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, \(f(x)=x^4\)

- 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? - Which differentiation rules are needed for an exponential with a nonconstant exponent? - After differentiating, which parts of the expression could affect its sign? - How can you confirm that concavity actually changes at each candidate?

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 \((\mu-\sigma, \mu+\sigma)\).

- Find the function value and derivative at the stated inflection point. - Use point-slope form to write a tangent equation. - To find where the tangent meets the x-axis, set its output equal to zero.

1. At \(x=\mu+\sigma\), the function value is \(\varphi_{\mu,\sigma}(\mu+\sigma)=\frac{1}{\sigma\sqrt{2\pi e}}\). 2. Since \(\varphi_{\mu,\sigma}'(x)=-\frac{x-\mu}{\sigma^2}\varphi_{\mu,\sigma}(x)\), the slope is \(-\frac{1}{\sigma^2\sqrt{2\pi e}}\). 3. The tangent is \(t(x)=-\frac{1}{\sigma^2\sqrt{2\pi e}}(x-\mu-\sigma)+\frac{1}{\sigma\sqrt{2\pi e}}\). 4. Setting \(t(x)=0\) and 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.

- Translate “increasing most rapidly” into a statement about the first derivative. - Which derivative helps locate a maximum of the first derivative? - Use a sign test or the next derivative to verify the maximum.

1. Differentiate using the product and chain rules: \(v'(t)=(1.5-0.05t)e^{0.05t}\). 2. The sales rate is increasing most rapidly where \(v'\) reaches its maximum. Differentiate again: \(v''(t)=(0.025-0.0025t)e^{0.05t}\). 3. Set \(v''(t)=0\). Since the exponential factor is positive, \(0.025-0.0025t=0\), so \(t=10\). 4. The third derivative is \(v'''(t)=(-0.00125-0.000125t)e^{0.05t}\), and \(v'''(10)<0\). Therefore, \(v'\) has a maximum at \(t=10\).

The sales rate is increasing most rapidly \(10\) days after release.

- 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 \((2, \infty)\).

- 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 \((-\infty, -2)\)

- 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-point x-coordinate. 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 signs for sufficiently large negative and positive \(x\). Because it is continuous, its sign must change somewhere between those two regions. Therefore, at least one of its zeros is a sign-changing zero. 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.

- 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

- 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 \(\left(\frac{\ln(-a)}{1-a}, \frac{a^2-1}{a}(-a)^{\frac{1}{1-a}}\right)\).

- 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.

- Calculate the first two derivatives and substitute \(x=2\). - The equations \(g'(2)=0\) and \(g''(2)=0\) are not sufficient by themselves. - Check the sign of \(g'\) on each side of \(x=2\). - A very flat graph can still have a local minimum.

1. Differentiate: \(g'(x)=0.4x^3-2.4x^2+4.8x-3.2=0.4(x-2)^3\) and \(g''(x)=1.2x^2-4.8x+4.8=1.2(x-2)^2\). 2. Therefore, \(g'(2)=0\) and \(g''(2)=0\). 3. For \(x<2\), \(g'(x)=0.4(x-2)^3<0\). For \(x>2\), \(g'(x)>0\). Thus, \(g\) changes from decreasing to increasing at \(x=2\). 4. By the first derivative test, \(g\) has a local minimum at \(x=2\), not a stationary inflection point. Equivalently, \(g'''(2)=0\) and \(g^{(4)}(2)=2.4>0\), so the first nonzero higher derivative has even order and is positive.

a) \(g'(2)=0\) and \(g''(2)=0\) b) The claim is false. Since \(g'\) changes from negative to positive, \(g\) has a local minimum at \(x=2\), not a stationary inflection point.

- 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 at least 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 \(x\geq0\)