This problem set covers every topic from the Derivatives Applications lesson — critical points, optimization, related rates, L’Hôpital’s Rule, and tangent lines. Work through each problem on your own, then click Show Solution to check your work step by step.
📚 Table of Contents
- Critical Points & Increasing/Decreasing
- First & Second Derivative Tests
- Absolute Max & Min
- Optimization
- Related Rates
- L’Hôpital’s Rule
- Tangent Lines
- Mixed & Challenge Problems
1. Critical Points & Increasing/Decreasing
Problem 1. Find the critical points of \(f(x) = x^3 – 3x + 5\)
👁️ Show Solution
$$f'(x) = 3x^2 – 3 = 3(x^2 – 1) = 3(x-1)(x+1)$$
Critical points at \(x = -1\) and \(x = 1\)
Problem 2. Find the intervals where \(f(x) = x^4 – 8x^2\) is increasing and decreasing.
👁️ Show Solution
$$f'(x) = 4x^3 – 16x = 4x(x^2 – 4) = 4x(x-2)(x+2)$$
Critical points: \(x = -2, 0, 2\)
| Interval | f'(x) | Behavior |
|---|---|---|
| \((-\infty,-2)\) | − | Decreasing ↓ |
| \((-2, 0)\) | + | Increasing ↑ |
| \((0, 2)\) | − | Decreasing ↓ |
| \((2, \infty)\) | + | Increasing ↑ |
Problem 3. Find the critical points of \(f(x) = \frac{x^2}{x-1}\)
👁️ Show Solution
Use Quotient Rule:
$$f'(x) = \frac{2x(x-1) – x^2}{(x-1)^2} = \frac{x^2 – 2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}$$
Critical points: \(x = 0\) and \(x = 2\) (note: \(x = 1\) is not in the domain)
Problem 4. Find the intervals where \(f(x) = xe^{-x}\) is increasing and decreasing.
👁️ Show Solution
Product Rule:
$$f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 – x)$$
Since \(e^{-x} > 0\) always, sign depends on \((1-x)\):
- \(f'(x) > 0\) when \(x < 1\) → Increasing on \((-\infty, 1)\)
- \(f'(x) < 0\) when \(x > 1\) → Decreasing on \((1, \infty)\)
Critical point at \(x = 1\)
2. First & Second Derivative Tests
Problem 5. Classify the critical points of \(f(x) = x^3 – 3x + 5\) using the First Derivative Test.
👁️ Show Solution
From Problem 1: critical points at \(x = -1\) and \(x = 1\)
- \(x < -1\): \(f' > 0\) (increasing)
- \(-1 < x < 1\): \(f' < 0\) (decreasing)
- \(x > 1\): \(f’ > 0\) (increasing)
\(x = -1\): + to − → Local Max. \(f(-1) = 7\)
\(x = 1\): − to + → Local Min. \(f(1) = 3\)
Problem 6. Use the Second Derivative Test to classify the critical points of \(f(x) = 2x^3 – 3x^2 – 12x\)
👁️ Show Solution
$$f'(x) = 6x^2 – 6x – 12 = 6(x-2)(x+1)$$
Critical points: \(x = 2\) and \(x = -1\)
$$f”(x) = 12x – 6$$
- \(f”(2) = 18 > 0\) → Local Minimum at \((2, -20)\)
- \(f”(-1) = -18 < 0\) → Local Maximum at \((-1, 7)\)
Problem 7. Find the inflection points of \(f(x) = 2x^3 – 3x^2 – 12x\)
👁️ Show Solution
$$f”(x) = 12x – 6 = 0 \Rightarrow x = \frac{1}{2}$$
Sign change confirmed: concave down for \(x < \frac{1}{2}\), concave up for \(x > \frac{1}{2}\)
$$f\!\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) – 3\left(\frac{1}{4}\right) – 6 = -\frac{13}{2}$$
Inflection point at \(\left(\frac{1}{2},\ -\frac{13}{2}\right)\)
3. Absolute Max & Min
Problem 8. Find the absolute max and min of \(f(x) = x^2 – 4x + 1\) on \([0, 5]\)
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$$f'(x) = 2x – 4 = 0 \Rightarrow x = 2$$
| x | f(x) |
|---|---|
| 0 | 1 |
| 2 | −3 |
| 5 | 6 |
Absolute max: 6 at x = 5. Absolute min: −3 at x = 2.
Problem 9. Find the absolute max and min of \(f(x) = x^3 – 6x^2 + 9x + 2\) on \([0, 4]\)
👁️ Show Solution
$$f'(x) = 3x^2 – 12x + 9 = 3(x-1)(x-3) = 0 \Rightarrow x = 1,\ x = 3$$
| x | f(x) |
|---|---|
| 0 | 2 |
| 1 | 6 |
| 3 | 2 |
| 4 | 6 |
Absolute max: 6 at x = 1 and x = 4. Absolute min: 2 at x = 0 and x = 3.
Problem 10. Find the absolute max and min of \(f(x) = \frac{x}{x^2+1}\) on \([-2, 2]\)
👁️ Show Solution
$$f'(x) = \frac{(x^2+1) – x(2x)}{(x^2+1)^2} = \frac{1 – x^2}{(x^2+1)^2} = 0 \Rightarrow x = \pm 1$$
| x | f(x) |
|---|---|
| −2 | \(-\frac{2}{5}\) |
| −1 | \(-\frac{1}{2}\) |
| 1 | \(\frac{1}{2}\) |
| 2 | \(\frac{2}{5}\) |
Absolute max: \(\frac{1}{2}\) at x = 1. Absolute min: \(-\frac{1}{2}\) at x = −1.
4. Optimization
Problem 11. Find two positive numbers whose sum is 50 and whose product is maximized.
👁️ Show Solution
Constraint: \(x + y = 50 \Rightarrow y = 50 – x\)
$$P = x(50 – x) = 50x – x^2$$
$$P’ = 50 – 2x = 0 \Rightarrow x = 25, \quad y = 25$$
The two numbers are 25 and 25. Maximum product = 625.
Problem 12. A farmer wants to fence a rectangular area using a barn wall as one side. He has 120 ft of fence. What dimensions maximize the area?
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Three sides need fencing: \(x + 2y = 120 \Rightarrow x = 120 – 2y\)
$$A = xy = (120 – 2y)y = 120y – 2y^2$$
$$A’ = 120 – 4y = 0 \Rightarrow y = 30, \quad x = 60$$
60 ft × 30 ft gives maximum area of 1800 ft²
Problem 13. Find the dimensions of a rectangle with perimeter 60 cm that has the maximum area.
👁️ Show Solution
Constraint: \(2x + 2y = 60 \Rightarrow y = 30 – x\)
$$A = x(30 – x) = 30x – x^2$$
$$A’ = 30 – 2x = 0 \Rightarrow x = 15, \quad y = 15$$
15 cm × 15 cm. Maximum area = 225 cm²
Problem 14. A cylindrical can must hold 500 cm³. Find the radius and height that minimize the surface area.
👁️ Show Solution
Constraint: \(h = \frac{500}{\pi r^2}\)
$$SA = 2\pi r^2 + \frac{1000}{r}$$
$$\frac{dSA}{dr} = 4\pi r – \frac{1000}{r^2} = 0 \Rightarrow r^3 = \frac{250}{\pi} \Rightarrow r \approx 4.30 \text{ cm}$$
$$h \approx 8.60 \text{ cm}$$
r ≈ 4.30 cm, h ≈ 8.60 cm
Problem 15. A square’s side length is increasing at 3 cm/s. How fast is the area increasing when the side is 8 cm?
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$$A = s^2 \Rightarrow \frac{dA}{dt} = 2s\frac{ds}{dt} = 2(8)(3) = 48 \text{ cm}^2/\text{s}$$
Problem 16. Water drains from a conical tank (radius 4 m, height 8 m) at 2 m³/min. How fast is the water level dropping when the water is 4 m deep?
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Since \(\frac{r}{h} = \frac{1}{2}\), substitute \(r = \frac{h}{2}\):
$$V = \frac{\pi h^3}{12} \Rightarrow \frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$$
$$-2 = \frac{\pi(16)}{4}\frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{-1}{2\pi} \approx -0.159 \text{ m/min}$$
Problem 17. Two cars leave an intersection at the same time. Car A travels north at 60 mph, Car B travels east at 80 mph. How fast is the distance between them increasing after 1 hour?
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After 1 hour: \(x = 80\), \(y = 60\), \(z = 100\) miles
$$2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$$
$$100\frac{dz}{dt} = 80(80) + 60(60) = 10000 \Rightarrow \frac{dz}{dt} = 100 \text{ mph}$$
Problem 18. A street light is 15 ft tall. A 6 ft tall person walks away at 5 ft/s. How fast is the shadow lengthening?
👁️ Show Solution
Let \(x\) = distance from base, \(s\) = shadow length. By similar triangles:
$$\frac{15}{x + s} = \frac{6}{s} \Rightarrow 9s = 6x \Rightarrow s = \frac{2x}{3}$$
$$\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt} = \frac{2}{3}(5) = \frac{10}{3} \approx 3.33 \text{ ft/s}$$
6. L’Hôpital’s Rule
Problem 19. \(\displaystyle\lim_{x \to 0} \frac{x^2}{\sin(x)}\)
👁️ Show Solution
Gives \(\frac{0}{0}\). Apply L’Hôpital’s:
$$\lim_{x \to 0} \frac{2x}{\cos(x)} = \frac{0}{1} = 0$$
Problem 20. \(\displaystyle\lim_{x \to \infty} \frac{\ln(x)}{x}\)
👁️ Show Solution
Gives \(\frac{\infty}{\infty}\). Apply L’Hôpital’s:
$$\lim_{x \to \infty} \frac{1/x}{1} = 0$$
Problem 21. \(\displaystyle\lim_{x \to 0} \frac{1 – \cos(x)}{x^2}\)
👁️ Show Solution
Gives \(\frac{0}{0}\). Apply L’Hôpital’s:
$$\lim_{x \to 0} \frac{\sin(x)}{2x}$$
Still \(\frac{0}{0}\). Apply again:
$$\lim_{x \to 0} \frac{\cos(x)}{2} = \frac{1}{2}$$
Problem 22. \(\displaystyle\lim_{x \to 1} \frac{x^3 – 1}{x^2 – 1}\)
👁️ Show Solution
Gives \(\frac{0}{0}\). Apply L’Hôpital’s:
$$\lim_{x \to 1} \frac{3x^2}{2x} = \frac{3}{2}$$
Problem 23. \(\displaystyle\lim_{x \to \infty} \frac{3x^2 + 5}{e^x}\)
👁️ Show Solution
Gives \(\frac{\infty}{\infty}\). Apply L’Hôpital’s twice:
$$\lim_{x \to \infty} \frac{6x}{e^x} \Rightarrow \lim_{x \to \infty} \frac{6}{e^x} = 0$$
7. Tangent Lines
Problem 24. Find the tangent line to \(f(x) = x^3 – 4x\) at \(x = 2\)
👁️ Show Solution
\(f(2) = 0\) → point \((2, 0)\)
\(f'(x) = 3x^2 – 4 \Rightarrow f'(2) = 8\)
$$y = 8(x – 2) \Rightarrow y = 8x – 16$$
Problem 25. Find the tangent line to \(f(x) = \sqrt{x}\) at \(x = 9\)
👁️ Show Solution
\(f(9) = 3\) → point \((9, 3)\)
\(f'(x) = \frac{1}{2\sqrt{x}} \Rightarrow f'(9) = \frac{1}{6}\)
$$y – 3 = \frac{1}{6}(x – 9) \Rightarrow y = \frac{x}{6} + \frac{3}{2}$$
Problem 26. Find the tangent line to \(f(x) = e^{2x}\) at \(x = 0\)
👁️ Show Solution
\(f(0) = 1\) → point \((0, 1)\)
\(f'(x) = 2e^{2x} \Rightarrow f'(0) = 2\)
$$y = 2x + 1$$
8. Mixed & Challenge Problems 🔥
Problem 27. A particle moves along a line with position \(s(t) = t^3 – 6t^2 + 9t\). Find when the particle is at rest, and when it’s moving forward vs. backward.
👁️ Show Solution
$$v(t) = s'(t) = 3t^2 – 12t + 9 = 3(t-1)(t-3)$$
At rest: \(t = 1\) and \(t = 3\)
- \(0 < t < 1\): \(v > 0\) → moving forward
- \(1 < t < 3\): \(v < 0\) → moving backward
- \(t > 3\): \(v > 0\) → moving forward
Problem 28. Find the point on the curve \(y = \sqrt{x}\) closest to the point \((4, 0)\).
👁️ Show Solution
Minimize distance squared: \(D = (x-4)^2 + y^2 = (x-4)^2 + x\)
$$\frac{dD}{dx} = 2(x-4) + 1 = 0 \Rightarrow x = \frac{7}{2}$$
$$y = \sqrt{\frac{7}{2}} = \frac{\sqrt{14}}{2}$$
Closest point: \(\left(\frac{7}{2},\ \frac{\sqrt{14}}{2}\right)\)
Problem 29. \(\displaystyle\lim_{x \to 0} \frac{e^{3x} – 1}{x}\)
👁️ Show Solution
Gives \(\frac{0}{0}\). Apply L’Hôpital’s:
$$\lim_{x \to 0} \frac{3e^{3x}}{1} = 3$$
Problem 30. A 13-foot ladder leans against a wall. The bottom slides away at 5 ft/s. How fast is the top sliding down when the bottom is 5 feet from the wall?
👁️ Show Solution
\(x^2 + y^2 = 169\). When \(x = 5\): \(y = 12\)
$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$
$$2(5)(5) + 2(12)\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{25}{12} \approx -2.08 \text{ ft/s}$$
The top is sliding down at \(\frac{25}{12}\) ft/s.
Problem 31. Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius 5.
👁️ Show Solution
Let half-dimensions be \(x\) and \(y\). Constraint: \(x^2 + y^2 = 25 \Rightarrow y = \sqrt{25 – x^2}\)
$$A = 4xy = 4x\sqrt{25-x^2}$$
$$\frac{dA}{dx} = \frac{4(25 – 2x^2)}{\sqrt{25-x^2}} = 0 \Rightarrow x = \frac{5}{\sqrt{2}}, \quad y = \frac{5}{\sqrt{2}}$$
Dimensions: \(5\sqrt{2} \times 5\sqrt{2}\). It’s a square!
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