AP Calculus AB Concavity and the Second Derivative Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Concavity and the Second Derivative – Maths PDF

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This concavity practice solutions for AP Calculus AB covers concavity and the second derivative (Common Core AP CALC AB 5.7, AP CALC AB 5.8, AP CALC AB 5.9), with a free PDF to print.

This answer key gives a full worked solution for every problem in the concavity worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.

AP Calculus AB · Common Core State Standards: AP CALC AB 5.7, AP CALC AB 5.8, AP CALC AB 5.9

Concavity practice solutions: exercises 1 to 10

Exercise 1 – Concavity of a cubic

$f^{\prime}=3x^{2}-12x+9$ and $f^{\prime\prime}=6x-12$. Set $f^{\prime\prime}=0$: $x=2$.

For $x < 2$, $f^{\prime\prime} < 0$; for $x > 2$, $f^{\prime\prime} > 0$. $f(2)=8-24+18+1=3$.

Final answer: $\text{concave down on }(-\infty,2);\ \text{concave up on }(2,\infty);\ \text{inflection point }(2,3)$.

Exercise 2 – Concavity of a logarithm

$f^{\prime}=\frac{2x}{x^{2}+1}$ and $f^{\prime\prime}=\frac{2(x^{2}+1)-2x(2x)}{(x^{2}+1)^{2}}=\frac{2-2x^{2}}{(x^{2}+1)^{2}}$.

The denominator is positive, so the sign is that of $2(1-x)(1+x)$: negative for $|x| > 1$ and positive for $|x| < 1$. $f(\pm 1)=\ln 2$.

Final answer: $\text{concave up on }(-1,1);\ \text{concave down on }(-\infty,-1)\text{ and }(1,\infty);\ \text{inflections }(\pm 1,\ln 2)$.

Exercise 3 – Exponential inflection

$f^{\prime}=e^{-x}(1-x)$ and $f^{\prime\prime}=e^{-x}(x-2)$.

Zero at $x=2$, changing from negative to positive. $f(2)=2e^{-2}$.

Final answer: $\left(2,\ 2e^{-2}\right)$.

Exercise 4 – Always concave down

$f^{\prime\prime}=-\frac{1}{x^{2}} < 0$ for all $x > 0$, so the graph never changes concavity.

Final answer: $\text{concave down on }(0,\infty);\ \text{no inflection point}$.

Exercise 5 – Sine on an interval

$f^{\prime\prime}=-\sin x$. It is negative on $(0,\pi)$ and positive on $(\pi,2\pi)$.

Final answer: $\text{concave down on }(0,\pi);\ \text{concave up on }(\pi,2\pi);\ \text{inflection point }(\pi,0)$.

Exercise 6 – Inflection where f double prime is undefined

$f^{\prime\prime}=-\frac{2}{9}x^{-5/3}$. For $x < 0$, $x^{-5/3} < 0$ so $f^{\prime\prime} > 0$; for $x > 0$, $f^{\prime\prime} < 0$.

Final answer: $\text{concavity changes at }x=0:\ \text{inflection point }(0,0)$.

Exercise 7 – The bell curve

$f^{\prime}=-\frac{2x}{(x^{2}+1)^{2}}$ and $f^{\prime\prime}=\frac{6x^{2}-2}{(x^{2}+1)^{3}}$.

Set $6x^{2}-2=0$: $x=\pm\frac{1}{\sqrt{3}}$. The sign changes at both. $f\left(\pm\frac{1}{\sqrt{3}}\right)=\frac{3}{4}$.

Final answer: $\left(\pm\frac{1}{\sqrt{3}},\ \frac{3}{4}\right)$.

Exercise 8 – Zero but not an inflection

$f^{\prime\prime}=12x^{2}$ is zero at $0$ but positive on both sides.

Final answer: $\text{No: }f^{\prime\prime}\text{ does not change sign}$.

Exercise 9 – Second derivative test, cubic

$f^{\prime}=3x^{2}-12=0$ gives $x=\pm 2$. $f^{\prime\prime}=6x$.

$f^{\prime\prime}(2)=12 > 0$: relative minimum. $f^{\prime\prime}(-2)=-12 < 0$: relative maximum.

Final answer: $\text{max }f(-2)=19;\ \text{min }f(2)=-13$.

Maximum at x = −2, minimum at x = 2.

Exercise 10 – Second derivative test, exponential

$f^{\prime}=e^{-x}x(2-x)$: critical numbers $0$ and $2$. $f^{\prime\prime}=e^{-x}(x^{2}-4x+2)$.

$f^{\prime\prime}(0)=2 > 0$: minimum. $f^{\prime\prime}(2)=e^{-2}(4-8+2)=-2e^{-2} < 0$: maximum.

Final answer: $\text{min }f(0)=0;\ \text{max }f(2)=4e^{-2}$.

Exercises 11 and up

Exercise 11 – A test that fails

$f^{\prime}=4x^{3}$ is zero at $0$, and $f^{\prime\prime}(0)=12(0)=0$: inconclusive.

First Derivative Test: $f^{\prime} < 0$ for $x < 0$ and $f^{\prime} > 0$ for $x > 0$.

Final answer: $\text{relative (and absolute) minimum }f(0)=0$.

Exercise 12 – Concavity from the graph of f prime

f is concave up where $f^{\prime}$ is increasing. $f^{\prime\prime}=3x^{2}-3=3(x-1)(x+1)$ is positive for $|x| > 1$.

The graph of $f^{\prime}$ has a local maximum at $x=-1$ and a local minimum at $x=1$: $f^{\prime\prime}$ changes sign there.

Final answer: $\text{concave up on }(-\infty,-1)\text{ and }(1,\infty);\ \text{inflection points at }x=-1\text{ and }x=1$.

Exercise 13 – Conclusions from a sign table

Concave down where $f^{\prime\prime} < 0$, concave up where $f^{\prime\prime} > 0$. The sign changes at $x=0$ and $x=3$.

Final answer: $\text{down on }(-\infty,0)\text{ and }(3,\infty);\ \text{up on }(0,3);\ \text{inflections at }0\text{ and }3$.

Exercise 14 – All features of a cubic

$f^{\prime}=3x(x-2)$: increasing on $(-\infty,0)$ and $(2,\infty)$, decreasing on $(0,2)$. Relative max $f(0)=0$, relative min $f(2)=-4$.

$f^{\prime\prime}=6x-6$: concave down on $(-\infty,1)$, concave up on $(1,\infty)$, inflection point $(1,-2)$.

Final answer: $\text{max at }(0,0),\ \text{min at }(2,-4),\ \text{inflection at }(1,-2)$.

Maximum, minimum and inflection point.

Exercise 15 – Curve sketch

$f=x(x-3)^{2}$, so the zeros are $x=0$ and $x=3$. $f^{\prime}=3(x-1)(x-3)$, $f^{\prime\prime}=6x-12$.

$f^{\prime\prime}(1)=-6 < 0$: max $f(1)=4$. $f^{\prime\prime}(3)=6 > 0$: min $f(3)=0$. Inflection at $x=2$, $f(2)=2$.

Final answer: $\text{max }(1,4);\ \text{min }(3,0);\ \text{inflection }(2,2)$.

Exercise 16 – Finding coefficients

$f^{\prime\prime}=6x+2a$ is zero at $x=2$: $12+2a=0$, so $a=-6$.

$f^{\prime}(1)=3+2a+b=0$ gives $3-12+b=0$, so $b=9$.

Final answer: $a=-6,\ b=9$.

Exercise 17 – Growth of sales

$S^{\prime}=-3t^{2}+18t+20$ and $S^{\prime\prime}=-6t+18$. $S^{\prime\prime}=0$ at $t=3$, changing from positive to negative.

$S^{\prime}(3)=-27+54+20=47$.

Final answer: $t=3\text{ months;}\ \text{rate }47\text{ thousand dollars per month}$.

Exercise 18 – Logistic population

Growth is fastest at the inflection point. At that time $9e^{-t/2}=1$, so $e^{-t/2}=\frac{1}{9}$ and $t=2\ln 9\approx 4.39$.

Then $P=\frac{1000}{1+1}=500$.

Final answer: $t\approx 4.39\text{ years, population }500$.

Exercise 19 – Marginal cost

$C^{\prime}=0.03x^{2}-1.8x+30$ and $C^{\prime\prime}=0.06x-1.8$, zero at $x=30$ and changing from negative to positive.

$C^{\prime}(30)=27-54+30=3$.

Final answer: $x=30\text{ mugs, minimum marginal cost }\$3$.

Exercise 20 – Both tests together

$f^{\prime}=5x^{3}(x-4)$: critical numbers $0$ and $4$. $f^{\prime\prime}=20x^{3}-60x^{2}=20x^{2}(x-3)$.

$f^{\prime\prime}(4)=320 > 0$: relative minimum. $f^{\prime\prime}(0)=0$: inconclusive; use signs of $f^{\prime}$: $+$ for $x < 0$ and $-$ for $0 < x < 4$.

Final answer: $\text{relative max at }x=0\ (f=0);\ \text{relative min at }x=4\ (f=-256)$.

Exercise 21 – Justifying inflection points

$f^{\prime\prime}=0$ at $x=1$ and $x=-2$.

At $x=1$, $f^{\prime\prime}$ changes from negative to positive. At $x=-2$ the factor $(x+2)^{2}$ does not change sign ($f^{\prime\prime} < 0$ on both sides).

Final answer: $\text{inflection point only at }x=1$.

No sign change at x = -2.

Key facts behind this concavity practice solutions

Key facts

A positive f″ means concave up (slope increasing); a negative f″ means concave down (slope decreasing).
An inflection point needs a change in the sign of f″, not just f″ = 0.
Second Derivative Test: f′(c) = 0 and a positive f″(c) gives a relative minimum; a negative f″(c) gives a maximum.
If f″(c) = 0 the test fails; use the First Derivative Test.

How to use this concavity practice solutions

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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