Practice Problems – Applications of Derivatives – AP Calculus

Below are 20 practice problems covering all topics from Module 02B: Applications of Derivatives. Topics include: increasing/decreasing, concavity, optimization, related rates, linearization, L'Hôpital's Rule, and MVT. Each problem includes a hidden solution — click the button to reveal the step-by-step answer.

Problem 01 Increasing/Decreasing

Find the intervals on which $f(x) = x^3 - 3x$ is increasing and decreasing.

Solution:
$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$
Critical points: $x = -1$ and $x = 1$
On $(-\infty, -1)$: $f' > 0$ → increasing
On $(-1, 1)$: $f' < 0$ → decreasing
On $(1, \infty)$: $f' > 0$ → increasing

Problem 02 Local Extrema

Find the local maximum and minimum values of $f(x) = x^3 - 3x$.

Solution:
From Problem 1, $f'$ changes from $+$ to $-$ at $x = -1$ → local max at $x = -1$, $f(-1) = 2$
$f'$ changes from $-$ to $+$ at $x = 1$ → local min at $x = 1$, $f(1) = -2$

Problem 03 Concavity

Find the intervals of concavity for $f(x) = x^3 - 3x$.

Solution:
$f''(x) = 6x$
On $(-\infty, 0)$: $f'' < 0$ → concave down
On $(0, \infty)$: $f'' > 0$ → concave up
Inflection point at $x = 0$

Problem 04 Optimization

Find two positive numbers whose sum is 20 and whose product is maximized.

Solution:
Let $x$ and $y$ be the numbers. $x + y = 20$ ⇒ $y = 20 - x$
Product $P(x) = x(20 - x) = 20x - x^2$
$P'(x) = 20 - 2x = 0$ ⇒ $x = 10$, $y = 10$
$P''(x) = -2 < 0$ → maximum. Numbers are $10$ and $10$.

Problem 05 Related Rates

A spherical balloon is being inflated at 4 cm³/s. How fast is the radius increasing when the radius is 2 cm?

Solution:
$V = \frac{4}{3}\pi r^3$, $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
$4 = 4\pi (4) \frac{dr}{dt}$ ⇒ $\frac{dr}{dt} = \frac{4}{16\pi} = \frac{1}{4\pi} \approx 0.0796$ cm/s

Problem 06 Linearization

Find the linearization of $f(x) = \sqrt{x}$ at $x = 9$ and use it to approximate $\sqrt{9.1}$.

Solution:
$f(9) = 3$, $f'(x) = \frac{1}{2\sqrt{x}}$, $f'(9) = \frac{1}{6}$
$L(x) = 3 + \frac{1}{6}(x-9)$
$L(9.1) = 3 + \frac{1}{6}(0.1) = 3 + 0.01667 = 3.01667$
Actual $\sqrt{9.1} \approx 3.01662$

Problem 07 L'Hôpital's Rule

Find $\displaystyle \lim_{x \to 0} \frac{\sin 2x}{x}$.

Solution:
$\frac{0}{0}$ form. L'Hôpital: $\lim_{x \to 0} \frac{2\cos 2x}{1} = 2$

Problem 08 L'Hôpital's Rule

Find $\displaystyle \lim_{x \to \infty} \frac{\ln x}{x}$.

Solution:
$\frac{\infty}{\infty}$ form. L'Hôpital: $\lim_{x \to \infty} \frac{1/x}{1} = 0$

Problem 09 Mean Value Theorem

Find the value(s) of $c$ guaranteed by the MVT for $f(x) = x^2$ on $[0, 2]$.

Solution:
$\frac{f(2)-f(0)}{2-0} = \frac{4-0}{2} = 2$
$f'(c) = 2c = 2$ ⇒ $c = 1$

Problem 10 Optimization

A rectangular garden is to be fenced on three sides with 100 m of fencing. Find the dimensions that maximize the area.

Solution:
Let $x$ = width (fenced on two sides), $y$ = length (fenced on one side)
$2x + y = 100$ ⇒ $y = 100 - 2x$
Area $A = x(100 - 2x) = 100x - 2x^2$
$A'(x) = 100 - 4x = 0$ ⇒ $x = 25$, $y = 50$
Dimensions: $25$ m by $50$ m

Problem 11 Related Rates

A 10 ft ladder slides down a wall. If the bottom is pulled away at 2 ft/s, how fast is the top sliding down when the bottom is 6 ft from the wall?

Solution:
$x^2 + y^2 = 100$, $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
When $x = 6$, $y = 8$, $\frac{dx}{dt} = 2$
$2(6)(2) + 2(8)\frac{dy}{dt} = 0$ ⇒ $24 + 16\frac{dy}{dt} = 0$ ⇒ $\frac{dy}{dt} = -1.5$ ft/s

Problem 12 L'Hôpital's Rule

Find $\displaystyle \lim_{x \to 0^+} x \ln x$.

Solution:
Rewrite as $\frac{\ln x}{1/x}$ → $\frac{\infty}{\infty}$ form
L'Hôpital: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$

Problem 13 Concavity

Find the inflection points of $f(x) = x^4 - 6x^2$.

Solution:
$f'(x) = 4x^3 - 12x$, $f''(x) = 12x^2 - 12 = 12(x^2 - 1)$
$f''(x) = 0$ ⇒ $x = \pm 1$
Sign change: concavity changes at both → inflection points at $x = -1$ and $x = 1$

Problem 14 Error Estimation

The radius of a sphere is measured as 5 cm with a possible error of 0.1 cm. Estimate the maximum error in the volume.

Solution:
$V = \frac{4}{3}\pi r^3$, $dV = 4\pi r^2 dr$
$dr = 0.1$, $r = 5$ ⇒ $dV = 4\pi (25)(0.1) = 10\pi \approx 31.4$ cm³

Problem 15 Mean Value Theorem

Verify that the MVT applies to $f(x) = \sqrt{x}$ on $[0, 4]$ and find $c$.

Solution:
$f$ is continuous on $[0,4]$ and differentiable on $(0,4)$.
$\frac{f(4)-f(0)}{4-0} = \frac{2}{4} = \frac{1}{2}$
$f'(c) = \frac{1}{2\sqrt{c}} = \frac{1}{2}$ ⇒ $\sqrt{c} = 1$ ⇒ $c = 1$

Problem 16 L'Hôpital's Rule

Find $\displaystyle \lim_{x \to \infty} \frac{e^x}{x^2}$.

Solution:
$\frac{\infty}{\infty}$ form. L'Hôpital: $\lim_{x \to \infty} \frac{e^x}{2x}$ (still $\infty/\infty$)
Apply again: $\lim_{x \to \infty} \frac{e^x}{2} = \infty$

Problem 17 Optimization

Find the point on the parabola $y = x^2$ closest to the point $(0, 2)$.

Solution:
Distance squared: $d^2 = x^2 + (x^2 - 2)^2 = x^4 - 3x^2 + 4$
$f'(x) = 4x^3 - 6x = 2x(2x^2 - 3) = 0$ ⇒ $x = 0$ or $x = \pm\sqrt{1.5}$
$x = \pm\sqrt{1.5}$ gives smaller distance. Points: $(\pm\sqrt{1.5}, 1.5)$

Problem 18 Related Rates

Water is poured into a conical tank at 2 m³/min. The tank has height 6 m and radius 3 m. How fast is the water level rising when the water is 2 m deep?

Solution:
Similar triangles: $\frac{r}{h} = \frac{3}{6} = \frac{1}{2}$ ⇒ $r = \frac{h}{2}$
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\frac{h^2}{4}) h = \frac{\pi}{12} h^3$
$\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$ ⇒ $2 = \frac{\pi}{4} (4) \frac{dh}{dt}$ ⇒ $\frac{dh}{dt} = \frac{2}{\pi} \approx 0.637$ m/min

Problem 19 L'Hôpital's Rule

Find $\displaystyle \lim_{x \to \infty} \left(1 + \frac{2}{x}\right)^x$.

Solution:
Let $y = \left(1 + \frac{2}{x}\right)^x$, $\ln y = x \ln\left(1 + \frac{2}{x}\right)$
$\lim_{x \to \infty} \frac{\ln(1 + 2/x)}{1/x}$ → $\frac{0}{0}$
L'Hôpital: $\lim_{x \to \infty} \frac{\frac{-2/x^2}{1+2/x}}{-1/x^2} = \lim_{x \to \infty} \frac{2}{1+2/x} = 2$
$\ln y \to 2$ ⇒ $y \to e^2$

Problem 20 MVT Application

Show that the equation $x^3 + x - 1 = 0$ has exactly one real root.

Solution:
$f(0) = -1 < 0$, $f(1) = 1 > 0$ → by IVT, at least one root in $(0,1)$.
$f'(x) = 3x^2 + 1 > 0$ for all $x$, so $f$ is strictly increasing.
If there were two roots, MVT would imply $f'(c) = 0$ between them — contradiction.
Therefore exactly one real root.

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