AP Calculus AB Increasing, Decreasing and the First Derivative Test Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Increasing, Decreasing and the First Derivative Test – Maths PDF

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This first derivative answers for AP Calculus AB covers increasing, decreasing and the first derivative test (Common Core AP CALC AB 5.4, AP CALC AB 5.5, AP CALC AB 5.6), with a free PDF to print.

This answer key gives a full worked solution for every problem in the first derivative test 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 5.4, AP CALC AB 5.5, AP CALC AB 5.6

First derivative answers: exercises 1 to 10

Exercise 1 – Intervals for a parabola

$f^{\prime}=2x-6$, zero at $x=3$. For $x < 3$, $f^{\prime} < 0$; for $x > 3$, $f^{\prime} > 0$.

Final answer: $\text{decreasing on }(-\infty,3);\ \text{increasing on }(3,\infty)$.

Exercise 2 – Intervals for a cubic

$f^{\prime}=3x^{2}-6x-9=3(x+1)(x-3)$. Test values: $f^{\prime}(-2)=15 > 0$, $f^{\prime}(0)=-9 < 0$, $f^{\prime}(4)=15 > 0$.

Final answer: $\text{increasing on }(-\infty,-1)\text{ and }(3,\infty);\ \text{decreasing on }(-1,3)$.

Sign chart of f′.

Exercise 3 – First derivative test, quartic

$f^{\prime}=4x^{3}-16x=4x(x-2)(x+2)$. Critical numbers: $-2,0,2$.

Signs of $f^{\prime}$: $-$ on $(-\infty,-2)$, $+$ on $(-2,0)$, $-$ on $(0,2)$, $+$ on $(2,\infty)$.

Final answer: $\text{relative minima }f(\pm 2)=-16;\ \text{relative maximum }f(0)=0$.

Exercise 4 – First derivative test, exponential

$f^{\prime}=e^{-x}(1-x)$, zero at $x=1$. Positive for $x < 1$, negative for $x > 1$.

Final answer: $\text{relative maximum at }x=1,\ f(1)=\frac{1}{e}$.

Exercise 5 – A function with an asymptote

$f^{\prime}=\frac{2x(x-1)-x^{2}}{(x-1)^{2}}=\frac{x^{2}-2x}{(x-1)^{2}}=\frac{x(x-2)}{(x-1)^{2}}$. Critical numbers $0$ and $2$; $x=1$ is not in the domain.

Signs: $+$ on $(-\infty,0)$, $-$ on $(0,1)$ and $(1,2)$, $+$ on $(2,\infty)$.

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

Exercise 6 – A square root

$f^{\prime}=1-\frac{1}{\sqrt{x}}$, zero at $x=1$ (undefined at the endpoint $0$).

Negative on $(0,1)$, positive on $(1,\infty)$. $f(1)=1-2=-1$.

Final answer: $\text{relative (and absolute) minimum }-1\text{ at }x=1$.

Exercise 7 – A trigonometric function

$f^{\prime}=1-2\cos x=0$ gives $\cos x=\frac{1}{2}$: $x=\frac{\pi}{3}$ and $\frac{5\pi}{3}$.

Signs: $-$ on $\left(0,\frac{\pi}{3}\right)$, $+$ on $\left(\frac{\pi}{3},\frac{5\pi}{3}\right)$, $-$ on $\left(\frac{5\pi}{3},2\pi\right)$.

Final answer: $\text{relative min at }x=\frac{\pi}{3}\ (f=\frac{\pi}{3}-\sqrt{3});\ \text{relative max at }x=\frac{5\pi}{3}\ (f=\frac{5\pi}{3}+\sqrt{3})$.

Exercise 8 – x ln x

$f^{\prime}=\ln x+1$, zero at $x=\frac{1}{e}$. Negative for $x < \frac{1}{e}$, positive after.

Final answer: $\text{relative minimum }f\left(\frac{1}{e}\right)=-\frac{1}{e}$.

Exercise 9 – Reading a graph of the derivative

$f^{\prime} < 0$ for $x < -1$ and $f^{\prime} > 0$ for $x > -1$ (with $f^{\prime}(2)=0$ only at a point).

At $x=-1$ the sign changes from $-$ to $+$: relative minimum. At $x=2$ there is no sign change: no extremum.

Final answer: $\text{decreasing on }(-\infty,-1);\ \text{increasing on }(-1,\infty);\ \text{relative min at }x=-1\text{ only}$.

Exercise 10 – Critical but not extreme

$f^{\prime}=3x^{2}$ is zero at $x=0$ and positive for every other $x$.

There is no sign change at $0$, so $f$ is increasing on both sides.

Final answer: $\text{critical number }0,\ \text{no extremum}$.

Exercises 11 and up

Exercise 11 – Conclusions from a sign table

Increasing on $(-\infty,-2)$ and $(4,\infty)$; decreasing on $(-2,4)$.

At $-2$: $+$ to $-$, relative maximum. At $1$: $-$ to $-$, no extremum. At $4$: $-$ to $+$, relative minimum.

Final answer: $\text{rel. max at }-2;\ \text{no extremum at }1;\ \text{rel. min at }4$.

Exercise 12 – Relative versus absolute

$f^{\prime}=6(x-1)(x-2)$: $+,-,+$ around $1$ and $2$. Relative maximum $f(1)=5$, relative minimum $f(2)=4$.

Candidates for absolute extrema: $f(0)=0$, $f(1)=5$, $f(2)=4$, $f(3)=9$.

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

Exercise 13 – Minimum on an open interval

$f^{\prime}=1-\frac{1}{x^{2}}=\frac{(x-1)(x+1)}{x^{2}}$. On $(0,\infty)$, the only critical number is $x=1$.

$f^{\prime} < 0$ on $(0,1)$ and $f^{\prime} > 0$ on $(1,\infty)$, so $f$ decreases then increases.

Final answer: $\text{absolute minimum }f(1)=2$.

Exercise 14 – Finding coefficients

$f^{\prime}=3x^{2}+2ax+b$ must equal $3(x-1)(x-3)=3x^{2}-12x+9$.

Compare: $2a=-12$ and $b=9$.

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

Exercise 15 – Ball height

$s^{\prime}=-32t+64=-32(t-2)$. Positive for $t < 2$, negative for $t > 2$.

Final answer: $\text{rising on }(0,2),\ \text{falling on }(2,4);\ \text{max height }s(2)=64\text{ ft}$.

Exercise 16 – Average cost

$A^{\prime}=1-\frac{100}{x^{2}}=0$ gives $x=10$. Negative on $(0,10)$, positive on $(10,\infty)$.

$A(10)=10+10=20$.

Final answer: $10\text{ hoodies, minimum average cost }\$20$.

Exercise 17 – Hiker elevation

$E^{\prime}$ changes from $+$ to $-$ at $t=2$ and from $-$ to $+$ at $t=5$.

Final answer: $\text{local high point at }t=2\text{ h;}\ \text{local low point at }t=5\text{ h}$.

Exercise 18 – Profit on an interval

$P^{\prime}=-3x^{2}+18x+48=-3(x-8)(x+2)$. The only critical number in the interval is $x=8$.

Candidates: $P(0)=0$, $P(8)=448$, $P(10)=380$.

Final answer: $8\text{ backpacks, profit }\$448$.

Exercise 19 – Critical number with undefined derivative

$f^{\prime}=\frac{4}{3}x^{1/3}-\frac{4}{3}x^{-2/3}=\frac{4(x-1)}{3x^{2/3}}$. Critical numbers: $x=1$ (zero) and $x=0$ (undefined, but in the domain).

Since $x^{2/3} > 0$ for $x\neq 0$, the sign of $f^{\prime}$ is the sign of $x-1$: negative on $(-\infty,0)$ and $(0,1)$, positive on $(1,\infty)$.

Final answer: $\text{relative min }f(1)=-3\text{ at }x=1;\ \text{no extremum at }x=0$.

Vertical tangent at 0, minimum at 1.

Exercise 20 – Repeated factor

Critical numbers: $-1,0,3$. Signs: $+$ on $(-\infty,-1)$, $-$ on $(-1,0)$, $-$ on $(0,3)$, $+$ on $(3,\infty)$.

The factor $x^{2}$ never changes sign, so $x=0$ is not an extremum.

Final answer: $\text{increasing on }(-\infty,-1)\text{ and }(3,\infty);\ \text{rel. max at }-1;\ \text{rel. min at }3;\ \text{none at }0$.

No sign change at x = 0.

Key facts behind this first derivative answers

Key facts

A positive f′ means f is increasing; a negative f′ means f is decreasing.
Sign changes of f′ can occur only at critical numbers or where f′ is undefined.
First Derivative Test: + to − gives a relative maximum; − to + gives a relative minimum.
A critical number without a sign change, such as x = 0 for x³, is not an extremum.

How to use this first derivative 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.

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