# Mean Value Theorem Practice Problems with Answers and Solutions > 12 Mean Value Theorem practice problems with answers and full solutions: finding c, Rolle's theorem, checking hypotheses, bounds and counting roots. Source: https://practicecalculus.com/practice-problems/mean-value-theorem Updated: 2026-11-06 $$ f'(c) = \frac{f(b) - f(a)}{b - a} $$ ## At a glance - **Problems:** 12, with answers and full solutions - **Level:** Calculus 1 and AP Calculus AB, Unit 5 - **You need:** Derivatives and the meaning of [continuity](/practice-problems/continuity) - **Main idea:** Somewhere between a and b, the tangent slope equals the average slope - **Practice with new numbers:** [Linear approximation and MVT lesson](/learn/calculus-1/applications-of-derivatives/linear-approximation-and-mvt) ## Key takeaways - Check both hypotheses first: $f$ continuous on $[a, b]$ and differentiable on $(a, b)$. - To find $c$, compute the average slope, set $f'(c)$ equal to it and keep only the solutions strictly between $a$ and $b$. - The MVT turns information about $f'$ into information about $f$. That is how you bound values and show an equation has only one root. 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 $(a, b)$, $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ See the [Mean Value Theorem](/theorems/mean-value-theorem) and [Rolle's theorem](/theorems/rolles-theorem) pages for the proof, or the [MVT guide](/blog/mean-value-theorem) 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 | Problem | Answer | | --- | --- | | 1 | $c = 2$ | | 2 | $c = \sqrt{3} \approx 1.732$ | | 3 | $c = 1$ | | 4 | $c = 2$ | | 5 | $c = e - 1 \approx 1.718$ | | 6 | $c = \pm\dfrac{2}{\sqrt{3}} \approx \pm 1.155$ | | 7 | $f(0) = f(4) = 1$, and $c = 2$ | | 8 | No, $f$ is not continuous at $x = 0$ | | 9 | No, $f$ is not differentiable at $0$, and no $c$ works | | 10 | By the MVT, the speed equals the average speed $75$ mph at some time | | 11 | $18$ | | 12 | The IVT gives at least one root, and Rolle's theorem rules out a second | ## Step-by-step solutions **1.** $f$ is a polynomial, so the hypotheses hold. The average slope is $\frac{9 - 1}{3 - 1} = 4$, and $f'(c) = 2c$: $$2c = 4 \quad\Rightarrow\quad c = 2$$ **2.** The average slope is $\frac{27 - 0}{3} = 9$, and $f'(c) = 3c^2$: $$3c^2 = 9 \quad\Rightarrow\quad c = \pm\sqrt{3}$$ 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 $\frac{2 - 0}{4} = \frac{1}{2}$: $$\frac{1}{2\sqrt{c}} = \frac{1}{2} \quad\Rightarrow\quad \sqrt{c} = 1 \quad\Rightarrow\quad c = 1$$ **4.** $\frac{1}{x}$ is fine on $[1, 4]$ because $0$ is not in the interval. The average slope is $\frac{\frac{1}{4} - 1}{3} = -\frac{1}{4}$: $$-\frac{1}{c^2} = -\frac{1}{4} \quad\Rightarrow\quad c^2 = 4 \quad\Rightarrow\quad c = 2$$ **5.** The average slope is $\frac{\ln e - \ln 1}{e - 1} = \frac{1}{e - 1}$, and $f'(c) = \frac{1}{c}$: $$\frac{1}{c} = \frac{1}{e - 1} \quad\Rightarrow\quad c = e - 1 \approx 1.718$$ 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 $\frac{2 - (-2)}{4} = 1$: $$3c^2 - 3 = 1 \quad\Rightarrow\quad c^2 = \frac{4}{3} \quad\Rightarrow\quad c = \pm\frac{2}{\sqrt{3}} \approx \pm 1.155$$ 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 $f'(c) = 0$: $$2c - 4 = 0 \quad\Rightarrow\quad c = 2$$ **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 $$s'(c) = \frac{s(2) - s(0)}{2 - 0} = \frac{150}{2} = 75 \text{ mph}$$ **11.** Apply the MVT on $[1, 4]$. For some $c$ in $(1, 4)$: $$f(4) - f(1) = f'(c)(4 - 1) \le 5\cdot 3 = 15$$ 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](/theorems/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](/learn/calculus-1/applications-of-derivatives/linear-approximation-and-mvt). To review the continuity hypothesis, use the [continuity practice problems](/practice-problems/continuity), and for a related idea with integrals, see the [Mean Value Theorem for integrals](/theorems/mean-value-theorem-for-integrals). ## Frequently asked questions ### What does the Mean Value Theorem say? If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there is at least one $c$ in $(a, b)$ with $f'(c) = \frac{f(b) - f(a)}{b - a}$. Some tangent line is parallel to the secant line through the endpoints. ### How do you find c in the Mean Value Theorem? Compute the average slope $\frac{f(b) - f(a)}{b - a}$, set $f'(c)$ equal to it and solve for $c$. Keep only the values strictly between $a$ and $b$. There can be more than one, as in problem 6. ### What is the difference between Rolle's theorem and the Mean Value Theorem? Rolle's theorem is the special case where $f(a) = f(b)$. Then the average slope is $0$, so there is a $c$ with $f'(c) = 0$, a horizontal tangent. The MVT is Rolle's theorem tilted.