Mean Value Theorem

If you drive $120$ km in $1.5$ hours, your average speed is $80$ km/h. The Mean Value Theorem says that at some instant your speedometer must have read exactly $80$ km/h. In calculus language: somewhere, the instantaneous rate of change equals the average rate of change. This theorem is the bridge between a function’s derivative and its overall behaviour, and it powers the rest of this unit.

Key ideas

The theorem

Mean Value Theorem (MVT). If $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there is at least one number $c$ in $(a, b)$ with

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

The right side is the slope of the secant line through $(a, f(a))$ and $(b, f(b))$: the average rate of change. The left side is the slope of the tangent line at $x = c$. So, geometrically, there is a point between $a$ and $b$ where the tangent line is parallel to the secant line.

For $f(x) = x^3 - x$ on $[0, 2]$, the tangent at $c = \frac{2}{\sqrt{3}} \approx 1.155$ is parallel to the secant (both have slope $3$).

The hypotheses matter

Both conditions are needed:

• Continuous on $[a, b]$: no holes, jumps, or asymptotes, including at the endpoints.
• Differentiable on $(a, b)$: no corners, cusps, or vertical tangents strictly between $a$ and $b$. (The endpoints don’t need to be differentiable.)

If either fails, the conclusion might fail. For example, $f(x) = \lvert x \rvert$ on $[-1, 2]$ has average rate of change $\frac{2 - 1}{3} = \frac{1}{3}$, but $f'(x)$ is only ever $-1$ or $1$. The corner at $x = 0$ breaks the theorem.

Polynomials, $e^x$, $\sin x$, and $\cos x$ are continuous and differentiable everywhere, so they always qualify. Rational functions and roots qualify on any interval that stays inside their domain (and, for roots, away from points with vertical tangents).

Rolle’s theorem: the special case

Rolle’s theorem. If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there is at least one $c$ in $(a, b)$ with $f'(c) = 0$.

This is just the MVT with a secant slope of $0$: if a smooth curve starts and ends at the same height, it must have a horizontal tangent somewhere in between.

What the theorem does and doesn’t tell you

The MVT guarantees that a $c$ exists. It doesn’t say where $c$ is, and there may be more than one. To actually find $c$, solve $f'(c) = \frac{f(b) - f(a)}{b - a}$ and keep only the solutions strictly between $a$ and $b$.

Justifying on the AP exam

On free-response questions, you usually apply the MVT to a function given by a table or in context. A full justification has three parts:

1. State the hypotheses are met: ”$f$ is differentiable, so it is continuous on $[a, b]$ and differentiable on $(a, b)$.”
2. Compute the average rate of change with the actual numbers.
3. Conclude with the theorem’s name: “By the Mean Value Theorem, there is a $c$ in $(a, b)$ with $f'(c) = \dots$”

A function that is differentiable is automatically continuous (see differentiability), so “differentiable” covers both hypotheses.

Worked examples

Example 1: Finding c for a quadratic

Show that $f(x) = x^2 + 2x$ satisfies the MVT hypotheses on $[1, 4]$, and find the value of $c$.

Solution. $f$ is a polynomial, so it is continuous on $[1, 4]$ and differentiable on $(1, 4)$.

The average rate of change is

$$\frac{f(4) - f(1)}{4 - 1} = \frac{24 - 3}{3} = 7$$

Now solve $f'(c) = 7$. Since $f'(x) = 2x + 2$:

$$2c + 2 = 7 \quad\Rightarrow\quad c = \frac{5}{2}$$

Check: $c = 2.5$ is in $(1, 4)$. ✓

Example 2: A cubic, keeping only c in the interval

Find all values of $c$ that satisfy the MVT for $f(x) = x^3 - x$ on $[0, 2]$.

Solution. $f$ is a polynomial, so the hypotheses are met. The average rate of change is

$$\frac{f(2) - f(0)}{2 - 0} = \frac{6 - 0}{2} = 3$$

Solve $f'(c) = 3c^2 - 1 = 3$:

$$3c^2 = 4 \quad\Rightarrow\quad c = \pm\frac{2}{\sqrt{3}} \approx \pm 1.155$$

Only $c = \dfrac{2}{\sqrt{3}} \approx 1.155$ is in $(0, 2)$, so that is the only value. (This is the point in the figure above.)

Example 3: When the hypotheses fail

Does the MVT apply to $g(x) = \dfrac{1}{x}$ on $[-1, 1]$? If you try anyway, what happens?

Solution. No. $g$ is not defined at $x = 0$, which is inside $[-1, 1]$, so $g$ is not continuous on the interval.

Trying anyway: the average rate of change is $\dfrac{g(1) - g(-1)}{1 - (-1)} = \dfrac{1 - (-1)}{2} = 1$. But $g'(x) = -\dfrac{1}{x^2}$ is negative for every $x$, so $g'(c) = 1$ has no solution. The conclusion fails because the hypotheses fail.

Example 4: A table, AP style

A cyclist rides along a straight path. Her distance from the start, $d(t)$ metres, is a differentiable function of time $t$ seconds. Some values:

$t$ (s)	$0$	$10$	$25$	$40$
$d(t)$ (m)	$0$	$80$	$230$	$380$

Must there be a time $t$ with $10 \lt t \lt 25$ when $d'(t) = 10$ m/s? Justify your answer.

Solution. Yes.

$d$ is differentiable, so it is continuous on $[10, 25]$ and differentiable on $(10, 25)$. The average rate of change on this interval is

$$\frac{d(25) - d(10)}{25 - 10} = \frac{230 - 80}{15} = 10 \text{ m/s}$$

By the Mean Value Theorem, there is a time $t$ in $(10, 25)$ with $d'(t) = 10$ m/s.

Common mistakes

Skipping the hypotheses. AP graders give a point for stating that the function is continuous and differentiable on the right intervals (or “differentiable, so also continuous”). Without it, the conclusion doesn’t follow.

Keeping a c outside the interval. In Example 2, $c = -\frac{2}{\sqrt{3}}$ solves the equation but isn’t in $(0, 2)$. Always check that $c$ is strictly between $a$ and $b$.

Using the MVT on a function with a break or a corner. Look for division by zero, absolute values, piecewise pieces, and roots with vertical tangents. If the function isn’t continuous on $[a, b]$ and differentiable on $(a, b)$, the theorem says nothing.

Mixing up the MVT with the IVT. The Intermediate Value Theorem is about function values (if $f$ is continuous, $f$ takes every value between $f(a)$ and $f(b)$). The MVT is about derivatives ($f'$ equals the average rate of change somewhere). Pick the theorem that matches what the question asks about.

Computing the average rate of change on the wrong interval. With a table, use exactly the two times the question names, not the first and last rows.

Thinking the MVT tells you where c is. It only says $c$ exists. If the function is given by a table, you usually can’t find $c$, and you don’t need to.

Practice

1. (Warm-up) Find the value of $c$ that satisfies the MVT for $f(x) = x^2$ on $[1, 3]$.

Solution

$f$ is a polynomial, so the hypotheses are met. The average rate of change is $\dfrac{9 - 1}{3 - 1} = 4$.

Solve $f'(c) = 2c = 4$, so $c = 2$, which is in $(1, 3)$.

2. (Warm-up) Explain why the MVT does not apply to $f(x) = \dfrac{1}{x}$ on $[-1, 2]$.

Solution

$f$ is undefined at $x = 0$, which is in $[-1, 2]$. So $f$ is not continuous on $[-1, 2]$, and the hypotheses are not met.

3. (Warm-up) Verify that Rolle’s theorem applies to $f(x) = x^2 - 4x$ on $[0, 4]$, and find $c$.

Solution

$f$ is a polynomial, so it is continuous and differentiable everywhere. Also $f(0) = 0$ and $f(4) = 16 - 16 = 0$, so $f(0) = f(4)$.

Solve $f'(c) = 2c - 4 = 0$: $c = 2$, which is in $(0, 4)$.

4. (Core) Find the value of $c$ that satisfies the MVT for $f(x) = \sqrt{x}$ on $[0, 9]$. Explain why the hypotheses are met.

Solution

$\sqrt{x}$ is continuous on $[0, 9]$, and $f'(x) = \dfrac{1}{2\sqrt{x}}$ exists for every $x$ in $(0, 9)$. (It’s not differentiable at $x = 0$, but the endpoint doesn’t need to be.)

The average rate of change is $\dfrac{3 - 0}{9 - 0} = \dfrac{1}{3}$.

$$\frac{1}{2\sqrt{c}} = \frac{1}{3} \quad\Rightarrow\quad \sqrt{c} = \frac{3}{2} \quad\Rightarrow\quad c = \frac{9}{4}$$

$c = 2.25$ is in $(0, 9)$.

5. (Core) Show that Rolle’s theorem applies to $f(x) = x^3 - 6x^2 + 9x$ on $[0, 3]$, and find every value of $c$ it guarantees.

Solution

$f$ is a polynomial, and $f(0) = 0$, $f(3) = 27 - 54 + 27 = 0$. So the hypotheses hold.

$$f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$$

$f'(c) = 0$ gives $c = 1$ or $c = 3$. Only $c = 1$ is in the open interval $(0, 3)$, so $c = 1$.

6. (Core) Find the value of $c$ that satisfies the MVT for $f(x) = \sin x$ on $\left[0, \dfrac{\pi}{2}\right]$. Give your answer to three decimal places. (Calculator allowed; use radians.)

Solution

$\sin x$ is continuous and differentiable everywhere. The average rate of change is

$$\frac{\sin\frac{\pi}{2} - \sin 0}{\frac{\pi}{2} - 0} = \frac{1}{\pi/2} = \frac{2}{\pi}$$

Solve $\cos c = \dfrac{2}{\pi}$ with $c$ in $\left(0, \dfrac{\pi}{2}\right)$:

$$c = \cos^{-1}\left(\frac{2}{\pi}\right) \approx 0.881$$

Make sure your calculator is in radian mode.

7. (Core) An oven heats up after being switched on. Its temperature $T(t)$, in °C, is a differentiable function of time $t$ in minutes.

$t$ (min)	$0$	$4$	$10$	$15$
$T(t)$ (°C)	$20$	$80$	$170$	$200$

Must there be a time $t$ with $4 \lt t \lt 10$ when $T'(t) = 15$ °C/min? Justify your answer.

Solution

Yes. $T$ is differentiable, so it is continuous on $[4, 10]$ and differentiable on $(4, 10)$. The average rate of change is

$$\frac{T(10) - T(4)}{10 - 4} = \frac{170 - 80}{6} = 15 \text{ °C/min}$$

By the Mean Value Theorem, there is a time $t$ in $(4, 10)$ with $T'(t) = 15$ °C/min.

8. (Challenge) Let $f(x) = x^2$ for $x \le 1$ and $f(x) = 2x - 1$ for $x \gt 1$. Show that the MVT applies on $[0, 3]$, and find $c$.

Solution

Continuity at $x = 1$: both pieces give $1$ there ($1^2 = 1$ and $2(1) - 1 = 1$), so $f$ is continuous. Each piece is a polynomial, so $f$ is continuous on $[0, 3]$.

Differentiability at $x = 1$: the left piece has derivative $2x$, which is $2$ at $x = 1$; the right piece has derivative $2$. The slopes match, so $f$ is differentiable at $1$, and therefore on $(0, 3)$.

The average rate of change is $\dfrac{f(3) - f(0)}{3} = \dfrac{5 - 0}{3} = \dfrac{5}{3}$.

On $(1, 3)$, $f'(x) = 2 \ne \frac{5}{3}$. On $(0, 1)$, solve $2c = \dfrac{5}{3}$: $c = \dfrac{5}{6}$, which is in $(0, 1)$. So $c = \dfrac{5}{6}$.

9. (Challenge) Show that $f(x) = x^3 + 2x - 5$ has exactly one real zero. (Hint: use the IVT to show there is one, then Rolle’s theorem to show there can’t be two.)

Solution

At least one: $f$ is continuous, $f(1) = 1 + 2 - 5 = -2 \lt 0$, and $f(2) = 8 + 4 - 5 = 7 \gt 0$. By the IVT, $f$ has a zero between $1$ and $2$.

At most one: suppose $f$ had two zeros, $a \lt b$. Then $f(a) = f(b) = 0$, and $f$ is a polynomial, so Rolle’s theorem gives a $c$ in $(a, b)$ with $f'(c) = 0$. But $f'(x) = 3x^2 + 2 \ge 2$ for every $x$, so $f'$ is never $0$. That’s a contradiction, so there can’t be two zeros.

So $f$ has exactly one real zero.