Mean Value Theorem Practice Problems with Answers and Solutions

These Mean Value Theorem practice problems start with finding the value of c, then cover Rolle's theorem, cases where the theorem does not apply, and the two classic uses: bounding a function and showing an equation has exactly one root. Try each one first. The answers come right after the list, then the full solutions.

The theorem: if f is continuous on [a, b] and differentiable on (a, b), then for some c in (a, b),

See the Mean Value Theorem and Rolle's theorem pages for the proof, or the MVT guide for more examples.

Problems

Find c

1. f(x) = x^2 on [1, 3]
2. f(x) = x^3 on [0, 3]
3. f(x) = \sqrt{x} on [0, 4]
4. f(x) = \dfrac{1}{x} on [1, 4]
5. f(x) = \ln x on [1, e]
6. f(x) = x^3 - 3x on [-2, 2]

Rolle's theorem

7. Verify Rolle's theorem for f(x) = x^2 - 4x + 1 on [0, 4] and find c.

Does the theorem apply?

8. Does the MVT apply to f(x) = \dfrac{1}{x} on [-1, 1]?
9. Does the MVT apply to f(x) = |x| on [-1, 2]? Is there a c that works anyway?

Using the theorem

10. A car travels 150 miles in 2 hours. Show that at some moment its speed was exactly 75 mph.
11. Suppose f(1) = 3 and f'(x) \le 5 for all x. What is the largest possible value of f(4)?
12. Show that x^3 + 2x - 1 = 0 has exactly one real solution.

Answers

Step-by-step solutions

1. f is a polynomial, so the hypotheses hold. The average slope is \frac{9 - 1}{3 - 1} = 4, and f'(c) = 2c:

2. The average slope is \frac{27 - 0}{3} = 9, and f'(c) = 3c^2:

Only c = \sqrt{3} is in (0, 3).

3. \sqrt{x} is continuous on [0, 4] and differentiable on (0, 4). That is enough, even though the derivative does not exist at x = 0. The average slope is \frac{2 - 0}{4} = \frac{1}{2}:

4. \frac{1}{x} is fine on [1, 4] because 0 is not in the interval. The average slope is \frac{\frac{1}{4} - 1}{3} = -\frac{1}{4}:

5. The average slope is \frac{\ln e - \ln 1}{e - 1} = \frac{1}{e - 1}, and f'(c) = \frac{1}{c}:

This is between 1 and e \approx 2.718, as it should be.

6. f(2) = 8 - 6 = 2 and f(-2) = -8 + 6 = -2, so the average slope is \frac{2 - (-2)}{4} = 1:

Both values are in (-2, 2), so both work. The MVT promises at least one c, not exactly one.

7. f is a polynomial, and f(0) = 1 and f(4) = 16 - 16 + 1 = 1 are equal, so Rolle's theorem applies. Solve f'(c) = 0:

8. No. \frac{1}{x} is undefined at x = 0, which is inside [-1, 1], so f is not continuous on the interval. In fact the average slope is \frac{1 - (-1)}{2} = 1, but f'(x) = -\frac{1}{x^2} is always negative, so no c works.

9. |x| is continuous on [-1, 2] but not differentiable at x = 0, so the MVT does not apply. The average slope is \frac{2 - 1}{3} = \frac{1}{3}, but f'(x) is only ever -1 or 1. No c gives \frac{1}{3}, which shows why the differentiability hypothesis matters.

10. Let s(t) be the distance traveled after t hours. Position is continuous and differentiable, with s'(t) the speed. By the MVT there is a time c in (0, 2) with

11. Apply the MVT on [1, 4]. For some c in (1, 4):

So f(4) \le 3 + 15 = 18. The value 18 happens when f'(x) = 5 everywhere, for example f(x) = 5x - 2.

12. At least one root. Let f(x) = x^3 + 2x - 1. It is continuous, with f(0) = -1 < 0 and f(1) = 2 > 0. By the Intermediate Value Theorem, there is a root in (0, 1).

At most one root. Suppose there were two roots a < b. Then f(a) = f(b) = 0, and Rolle's theorem would give a c with f'(c) = 0. But f'(x) = 3x^2 + 2 \ge 2 is never 0. So there cannot be two roots, and there is exactly one.

Common mistakes on these problems

• Skipping the hypotheses. Problems 8 and 9 show that without continuity or differentiability, the conclusion can fail.
• Keeping a c outside the interval. In problem 2, c = -\sqrt{3} solves the equation but is not in (0, 3).
• Stopping at one c. In problem 6 there are two valid values. List every solution in the interval.
• Mixing up average and instant rates. The left side of the MVT is a derivative at one point. The right side is the slope of the secant line over the whole interval.

Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new functions and instant feedback, try the linear approximation and MVT lesson. To review the continuity hypothesis, use the continuity practice problems, and for a related idea with integrals, see the Mean Value Theorem for integrals.