AP Calculus AB Mean Value Theorem and Extreme Value Theorem Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Mean Value Theorem and Extreme Value Theorem – Maths PDF

Cover of the book AP Calculus AB Math Worksheets with Answers (PDF)

Download as a PDFThe AP Calculus AB worksheets book with answersPrintable PDFSee the book ›

Sommaire

This mvt evt answers for AP Calculus AB covers mean value theorem and extreme value theorem (Common Core AP CALC AB 5.1, AP CALC AB 5.2, AP CALC AB 5.3), with a free PDF to print.

This answer key gives a full worked solution for every problem in the mean value theorem 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 5.1, AP CALC AB 5.2, AP CALC AB 5.3

Mvt evt answers: exercises 1 to 10

Exercise 1 – MVT with a parabola

$f$ is a polynomial, so it is continuous and differentiable everywhere.

Average rate: $\frac{16-1}{4-1}=5$. Solve $f^{\prime}(c)=2c=5$: $c=2.5$, which is in $(1,4)$.

Final answer: $c=\frac{5}{2}$.

Exercise 2 – MVT with a cubic

Average rate: $\frac{8-0}{2}=4$. Solve $3c^{2}=4$: $c=\pm\frac{2}{\sqrt{3}}$.

Only $\frac{2}{\sqrt{3}}\approx 1.155$ lies in $(0,2)$.

Final answer: $c=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$.

The tangent at c ≈ 1.155 is parallel to the secant.

Exercise 3 – MVT with a logarithm

Average rate: $\frac{\ln e^{2}-\ln 1}{e^{2}-1}=\frac{2}{e^{2}-1}$.

Solve $\frac{1}{c}=\frac{2}{e^{2}-1}$: $c=\frac{e^{2}-1}{2}\approx 3.19$, which lies in $(1,e^{2})$.

Final answer: $c=\frac{e^{2}-1}{2}$.

Exercise 4 – MVT with a reciprocal

Average rate: $\frac{\frac{1}{4}-1}{3}=-\frac{1}{4}$. Solve $-\frac{1}{c^{2}}=-\frac{1}{4}$: $c^{2}=4$, so $c=2$ (since $c > 0$).

Final answer: $c=2$.

Exercise 5 – Two values of c

$f(2)=2$ and $f(-2)=-2$, so the average rate is $\frac{2-(-2)}{4}=1$.

Solve $3c^{2}-3=1$: $c^{2}=\frac{4}{3}$, so $c=\pm\frac{2}{\sqrt{3}}$, both in $(-2,2)$.

Final answer: $c=\pm\frac{2\sqrt{3}}{3}$.

The tangent at c ≈ 1.155 (another is at −1.155).

Exercise 6 – When the MVT fails

The average rate is $\frac{2-1}{3}=\frac{1}{3}$. But $f$ is not differentiable at $x=0$, which is inside the interval, so the MVT does not apply.

Indeed $f^{\prime}(x)=-1$ for $x < 0$ and $1$ for $x > 0$, never $\frac{1}{3}$.

Final answer: $\text{No. The theorem does not apply, and no such }c\text{ exists}$.

Exercise 7 – Discontinuity

$f$ is undefined at $x=0$, so it is not continuous on $[-1,1]$. Also $f^{\prime}(x)=-\frac{1}{x^{2}} < 0$ everywhere it exists, while the average rate is $\frac{1-(-1)}{2}=1 > 0$.

Final answer: $\text{not continuous on }[-1,1]$.

Exercise 8 – Speeding ticket

Distance: $130-25=105$ miles in $1.5$ hours. Average speed: $\frac{105}{1.5}=70$ mph.

Position is continuous and differentiable, so by the MVT the speed was exactly $70$ mph at some instant, which exceeds 65.

Final answer: $70\text{ mph average, so speeding at some instant}$.

Exercise 9 – MVT from a table

Average rate: $\frac{11-3}{4}=2$. The MVT guarantees a $c$ in $(0,4)$ with $f^{\prime}(c)=2$.

Therefore $f^{\prime}(x)$ cannot be less than 2 everywhere.

Final answer: $f^{\prime}(c)=2\text{ for some }c;\ \text{no}$.

Exercise 10 – Rolle's theorem

$f(1)=0$ and $f(3)=0$, so the average rate is $0$. $f^{\prime}(c)=2c-4=0$ gives $c=2$.

Final answer: $c=2$.

Exercises 11 and up

Exercise 11 – EVT with a parabola

$f^{\prime}=2x=0$ gives the critical number $x=0$. Candidates: $f(-1)=1$, $f(0)=0$, $f(3)=9$.

Final answer: $\text{absolute maximum }9\text{ at }x=3;\ \text{absolute minimum }0\text{ at }x=0$.

Exercise 12 – Critical numbers with a Gaussian factor

Product and chain rules: $f^{\prime}(x)=e^{-x^{2}/2}+x\cdot e^{-x^{2}/2}(-x)=e^{-x^{2}/2}(1-x^{2})$.

The exponential is never zero, so solve $1-x^{2}=0$. The derivative exists everywhere.

Final answer: $x=-1\text{ and }x=1$.

Exercise 13 – Undefined derivative

$f^{\prime}(x)=\frac{2}{3}x^{-1/3}=\frac{2}{3\sqrt[3]{x}}$ is never zero, but is undefined at $x=0$, which is in the domain.

Final answer: $x=0$.

Exercise 14 – Product with an exponential

$f^{\prime}(x)=e^{-x}-xe^{-x}=e^{-x}(1-x)$. Since $e^{-x} > 0$, set $1-x=0$.

Final answer: $x=1$.

Exercise 15 – Trig critical numbers

$f^{\prime}=\cos x-\sin x=0$ means $\tan x=1$.

On $(0,2\pi)$: $x=\frac{\pi}{4}$ and $x=\frac{5\pi}{4}$.

Final answer: $x=\frac{\pi}{4},\ \frac{5\pi}{4}$.

Exercise 16 – Closed interval method, cubic

$f^{\prime}=3x^{2}-3=0$ gives $x=\pm 1$, both in the interval.

$f(-2)=-2$, $f(-1)=2$, $f(1)=-2$, $f(3)=18$.

Final answer: $\text{max }18\text{ at }x=3;\ \text{min }-2\text{ at }x=-2\text{ and }x=1$.

Exercise 17 – Closed interval method, second cubic

$f^{\prime}=3x^{2}-12x+9=3(x-1)(x-3)$. Candidates: $f(0)=1$, $f(1)=5$, $f(3)=1$, $f(5)=21$.

Final answer: $\text{max }21\text{ at }x=5;\ \text{min }1\text{ at }x=0\text{ and }x=3$.

The candidates on [0, 5].

Exercise 18 – Closed interval method, rational

$f^{\prime}=\frac{1-x^{2}}{(x^{2}+1)^{2}}$, zero at $x=1$. Candidates: $f(0)=0$, $f(1)=\frac{1}{2}$, $f(4)=\frac{4}{17}$.

Final answer: $\text{max }\frac{1}{2}\text{ at }x=1;\ \text{min }0\text{ at }x=0$.

Exercise 19 – Endpoint critical number

$f^{\prime}=\frac{1}{x}-1=0$ gives $x=1$, an endpoint. Candidates: $f(1)=-1$ and $f(4)=\ln 4-4\approx -2.614$.

Final answer: $\text{max }-1\text{ at }x=1;\ \text{min }\ln 4-4\text{ at }x=4$.

Exercise 20 – Greenhouse temperature

$T^{\prime}=-2t+12=0$ gives $t=6$. Candidates: $T(0)=40$, $T(6)=76$, $T(10)=60$.

Final answer: $\text{highest }76\,^{\circ}\text{F at }t=6;\ \text{lowest }40\,^{\circ}\text{F at }t=0$.

Exercise 21 – Maximum revenue

$R^{\prime}=120-2x=0$ gives $x=60$. Candidates: $R(0)=0$, $R(60)=3600$, $R(100)=2000$.

Final answer: $60\text{ boards, revenue }\$3600$.

Exercise 22 – Parallel tangent

Average rate: $\frac{6-0}{2}=3$. $3c^{2}-1=3$ gives $c=\frac{2}{\sqrt{3}}$.

$f(c)=\frac{8}{3\sqrt{3}}-\frac{2}{\sqrt{3}}=\frac{2}{3\sqrt{3}}$. Tangent: $y-\frac{2}{3\sqrt{3}}=3\left(x-\frac{2}{\sqrt{3}}\right)$.

Final answer: $c=\frac{2}{\sqrt{3}}\approx 1.155$.

Key facts behind this mvt evt answers

Key facts

MVT: f′(c) = (f(b) – f(a))/(b – a) for some c in (a, b), if f is continuous and differentiable.
EVT: a continuous function on a closed interval has an absolute maximum and an absolute minimum.
Critical numbers are where f′ = 0 or f′ is undefined (inside the domain).
Candidates test: compare f at endpoints and critical numbers.

How to use this mvt evt answers

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.

Download the mvt evt answers as a PDF with the button below and keep it with your notes.

All AP Calculus AB chapters: AP Calculus AB math.

Voter.. post

Free printable PDF