The Mean Value Theorem in one picture, and the two conditions it needs

By Marius, private math tutor in Montreal · 2026-10-05

The Mean Value Theorem says that if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then somewhere strictly between $a$ and $b$ the tangent line is parallel to the secant line through the two endpoints: $f^{'} (c)$ equals the average rate of change. Take away either condition and that guarantee is gone. Here is the picture, a counterexample for each condition, and a self-check with answers.

The statement, one piece at a time

The secant line joins the endpoints $(a, f (a))$ and $(b, f (b))$ . Its slope is the average rate of change of $f$ over the interval: $\frac{f (b) - f (a)}{b - a}$
The tangent line at a point $c$ has slope $f^{'} (c)$ , the instantaneous rate of change there.
The theorem. 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 $f^{'} (c) = \frac{f (b) - f (a)}{b - a}$

In words: at some moment, the instantaneous rate equals the average rate. A car that covers 120 km in 1.5 hours averages 80 km/h, so at some instant its speedometer read exactly 80 km/h.

The picture, worked through

Take $f (x) = 4 - \frac{4}{x}$ on $[1, 4]$ .

Check both conditions. $f$ is a rational function whose only break is at $x = 0$ , outside $[1, 4]$ , so it is continuous on $[1, 4]$ . Its derivative, $f^{'} (x) = \frac{4}{x^{2}}$ , exists at every $x \neq 0$ , so $f$ is differentiable on $(1, 4)$ .
Average rate of change. $f (1) = 0$ and $f (4) = 3$ , so $\frac{f (4) - f (1)}{4 - 1} = \frac{3 - 0}{3} = 1$
Find $c$ . Solve $\frac{4}{c^{2}} = 1$ : $c^{2} = 4$ , so $c = 2$ or $c = - 2$ . Only $c = 2$ lies in $(1, 4)$ .
Check the picture. At $c = 2$ the point is $(2, 2)$ , and the tangent there is $y = 2 + 1 \cdot (x - 2)$ , that is $y = x$ . The secant through $(1, 0)$ and $(4, 3)$ is $y = x - 1$ . Same slope, different lines: parallel.

The secant (solid) and the tangent at

c = 2

(dashed) both have slope

1

. Notice that

c = 2

is not the midpoint of

[1, 4]

: the theorem says that a

c

exists, not where it is.

Why continuity on the closed interval matters

Continuity has to hold at the endpoints too. Let $g (x) = x^{2}$ for $0 \leq x < 2$ , and set $g (2) = - 1$ .

$g$ is differentiable on $(0, 2)$ , with $g^{'} (x) = 2 x$ .
The average rate of change is $\frac{g (2) - g (0)}{2 - 0} = \frac{- 1 - 0}{2} = - \frac{1}{2}$ .
But $g^{'} (c) = 2 c$ is positive for every $c$ in $(0, 2)$ , so no $c$ works.

The only defect is at $x = 2$ : the values climb toward $4$ and the graph then lands at $- 1$ . One misplaced endpoint is enough to break the guarantee.

Why differentiability on the open interval matters

Let $p (x) = x^{2 / 3}$ on $[- 1, 1]$ . It is continuous everywhere, and $p (- 1) = p (1) = 1$ , so the average rate of change is $0$ : the secant is horizontal. A parallel tangent would need $p^{'} (c) = 0$ , but

p^{'} (x) = \frac{2}{3 \sqrt[3]{x}} (x \neq 0)

is never $0$ , and at $x = 0$ the graph has a cusp with no derivative at all. No $c$ exists: the graph turns around at the origin without ever being flat.

Left:

g

, which jumps at

x = 2

. Right:

p (x) = x^{2 / 3}

, which has a cusp at

0

. In both, no tangent line is parallel to the secant.

Why the endpoints only need continuity

The theorem asks for differentiability on the open interval, and that is deliberate. The upper half of a circle, $s (x) = \sqrt{4 - x^{2}}$ on $[- 2, 2]$ , has a vertical tangent at each endpoint, so it has no derivative at $x = \pm 2$ . It is still continuous on $[- 2, 2]$ and differentiable on $(- 2, 2)$ , so the theorem applies: the secant is horizontal, and the tangent at $c = 0$ , the top of the arc, is horizontal too.

The endpoints have to be attached to the curve, which is what continuity there means. How steep the curve is at $a$ or $b$ never matters.

When a condition fails, the theorem says nothing. The correct conclusion is “the Mean Value Theorem does not apply”: a suitable $c$ is no longer guaranteed. The trap is writing “so no such $c$ exists”. Sometimes one exists anyway, as question 5 below shows. To prove that none exists, you have to check the derivative directly, as in the two counterexamples above.

What the theorem is for

Written as $f (b) - f (a) = f^{'} (c) (b - a)$ , the theorem turns facts about the derivative into facts about the function. That is how results you use all year are proved:

If $f^{'} (x) > 0$ on an interval, then $f$ is increasing there: for any $a < b$ in the interval, $f (b) - f (a) = f^{'} (c) (b - a) > 0$ .
If $f^{'} (x) = 0$ on an interval, then $f$ is constant there.
A bound on $f^{'}$ is a bound on how far $f$ can move, which is question 6 below.

Using the theorem also means saying that it applies. The College Board's Course and Exam Description asks AP students to make a habit of “explicitly verifying that a theorem’s hypotheses are satisfied before applying the theorem.” Here that is one clause for continuity on $[a, b]$ and one for differentiability on $(a, b)$ .

Check your understanding

No calculator needed.

Find the value of $c$ that the Mean Value Theorem guarantees for $f (x) = \ln x$ on $[1, e]$ .
Find every value of $c$ that the theorem guarantees for $f (x) = x^{3} - 3 x^{2}$ on $[- 1, 4]$ .
Explain why the Mean Value Theorem does not apply to $f (x) = \frac{1}{x}$ on $[- 1, 1]$ , then show that no $c$ in $(- 1, 1)$ satisfies its conclusion.
Does the theorem apply to $g (x) = \sqrt[3]{x}$ on $[0, 8]$ , even though $g^{'} (0)$ does not exist? If it does, find $c$ .
True or false: let $h (x) = x^{3}$ for $0 < x \leq 2$ and $h (0) = 2$ . Because $h$ is not continuous on $[0, 2]$ , no $c$ in $(0, 2)$ has $h^{'} (c)$ equal to the average rate of change of $h$ over $[0, 2]$ .
A function $f$ is differentiable everywhere, $f (1) = 2$ , and $f^{'} (x) \leq 4$ for every $x$ . What is the largest value $f (6)$ can have?
Find the error: “A cyclist rides 60 km in 3 hours, so by the Mean Value Theorem she was riding at exactly 20 km/h at $t = 1.5$ h, halfway through the ride.”

Show the answers

$c = e - 1 \approx 1.72$ . The average rate is $\frac{\ln e - \ln 1}{e - 1} = \frac{1}{e - 1}$ , and $f^{'} (c) = \frac{1}{c}$ , so $c = e - 1$ , which lies in $(1, e)$ .
$c = 1 \pm \frac{\sqrt{21}}{3}$ , about $- 0.53$ and $2.53$ . Here $f (- 1) = - 4$ and $f (4) = 16$ , so the average rate is $\frac{20}{5} = 4$ ; then $3 c^{2} - 6 c = 4$ gives $c = 1 \pm \frac{\sqrt{21}}{3}$ . Both lie in $(- 1, 4)$ , so both count: the theorem promises at least one $c$ , not exactly one.
$f$ is undefined at $0$ , which is inside $[- 1, 1]$ , so it is not continuous on the interval and the theorem guarantees nothing. The average rate is $\frac{1 - (- 1)}{2} = 1$ , but $f^{'} (x) = - \frac{1}{x^{2}}$ is negative wherever it exists, so it never equals $1$ .
Yes, with $c = \frac{8 \sqrt{3}}{9} \approx 1.54$ . $g$ is continuous on $[0, 8]$ and differentiable on $(0, 8)$ ; the bad point $x = 0$ is an endpoint, where only continuity is needed. The average rate is $\frac{2 - 0}{8} = \frac{1}{4}$ , and $\frac{1}{3 c^{2 / 3}} = \frac{1}{4}$ gives $c^{2 / 3} = \frac{4}{3}$ , so $c = {(\frac{4}{3})}^{3 / 2}$ .
False. The average rate is $\frac{h (2) - h (0)}{2 - 0} = \frac{8 - 2}{2} = 3$ , and $h^{'} (c) = 3 c^{2} = 3$ at $c = 1$ , which is in $(0, 2)$ . The theorem does not apply, so it promises nothing, but its conclusion can still happen to hold.
$22$ . On $[1, 6]$ the theorem gives $f (6) - f (1) = f^{'} (c) \cdot 5 \leq 4 \cdot 5 = 20$ , so $f (6) \leq 22$ . The line $f (x) = 4 x - 2$ reaches it.
The theorem promises some instant in $(0, 3)$ at which her speed was 20 km/h, not that the instant is the halfway point. If her distance were $s (t) = 60 {(\frac{t}{3})}^{3}$ kilometres, she would be at 15 km/h at $t = 1.5$ and reach 20 km/h only at $t = \sqrt{3} \approx 1.73$ h.

For practice that asks you to check both conditions in writing, and to reject a $c$ that lands outside the open interval, work through the free AP Calculus AB Unit 5 worksheet. CEGEP students will find the same theorem, alongside Rolle's theorem, in the free CEGEP Calculus I curve-sketching worksheet.

Want one-on-one help with this? I offer AP Calculus AB tutoring and exam preparation in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.

Practise it on a free worksheet

Sources

AP Calculus AB and BC Course and Exam Description (Unit 1 overview; topic 5.1, Using the Mean Value Theorem) — College Board. Checked 2026-10-05.