The Mean Value Theorem, calculus lesson

Suppose you drive 150 kilometers in 2 hours. Your average speed is 75 kilometers per hour, and at some instant your speedometer must have read exactly 75. You may have gone slower and faster, but you cannot have been below 75 the whole time or above it the whole time. The mean value theorem turns this everyday reasoning into a precise statement, and much of the rest of calculus leans on it.

Key terms

Rolle’s theorem

Definition Let $a < b$ and let $f$ be a function with these three properties.

It is continuous on the closed interval $[a, b]$ .
It is differentiable on the open interval $(a, b)$ .
Its values at the ends agree, $f(a) = f(b)$ .

Then there is at least one number $c$ in $(a, b)$ with $f'(c) = 0$ .

The proof uses the extreme value theorem. If $f$ is constant, every $c$ in $(a, b)$ works. Otherwise some value of $f$ on $[a, b]$ is above or below the common end value $f(a) = f(b)$ . Then the global maximum or the global minimum, which exist by continuity, is attained at an interior point $c$ , and Fermat’s theorem gives $f'(c) = 0$ because $f$ is differentiable at $c$ .

In plain words If a smooth curve starts and ends at the same height, somewhere in between it has a horizontal tangent.

Example Let $f(x) = x^2 - 4x$ on $[0, 4]$ . It is a polynomial, so it is continuous and differentiable, and $f(0) = f(4) = 0$ . Solving $f'(c) = 2c - 4 = 0$ gives $c = 2$ , which lies in $(0, 4)$ .

Mean value theorem

Definition Let $a < b$ . If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then there is at least one number $c$ in $(a, b)$ with

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

It follows from Rolle’s theorem applied to $f$ minus its secant line,

g(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a),

which is continuous on $[a, b]$ , differentiable on $(a, b)$ and has $g(a) = g(b) = 0$ .

In plain words Somewhere between $a$ and $b$ the tangent line is parallel to the secant line joining the two ends. The instantaneous rate equals the average rate at least once.

Example Let $f(x) = x^3$ on $[0, 2]$ . The average rate is

\frac{f(2) - f(0)}{2 - 0} = \frac{8 - 0}{2 - 0} = 4.

Solving $3c^2 = 4$ gives $c = \frac{2}{\sqrt{3}} \approx 1.155$ , which is in $(0, 2)$ . The other solution, $-\frac{2}{\sqrt{3}}$ , is outside the interval.

The hypotheses

Definition The conditions continuity on $[a, b]$ and differentiability on $(a, b)$ are part of both theorems. Differentiability is not needed at the endpoints, and continuity is needed at every point including the endpoints. If either condition fails, the conclusion can fail. The conditions are sufficient, not necessary, so a function can break them and still have a suitable $c$ .

In plain words One corner or one jump is enough to break the theorem. It is not enough to check only that the function is defined.

Example The function $f(x) = \lvert x \rvert$ on $[-1, 1]$ has $f(-1) = f(1) = 1$ , yet $f'(x)$ is $-1$ or $1$ and never $0$ . Rolle’s theorem does not apply because $f$ is not differentiable at $0$ . The function $g(x) = x$ for $0 \le x < 1$ with $g(1) = 0$ has $g(0) = g(1)$ and $g'(x) = 1$ on $(0, 1)$ . Here continuity at the endpoint $1$ fails.

Cauchy’s mean value theorem

Definition Let $a < b$ , and let $f$ and $g$ be continuous on $[a, b]$ and differentiable on $(a, b)$ . Then there is a number $c$ in $(a, b)$ with

\big(f(b) - f(a)\big) g'(c) = \big(g(b) - g(a)\big) f'(c).

If in addition $g'(x) \neq 0$ for all $x$ in $(a, b)$ , then $g(b) \neq g(a)$ and

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

The proof applies Rolle’s theorem to

h(x) = \big(f(b) - f(a)\big) g(x) - \big(g(b) - g(a)\big) f(x),

which has $h(a) = h(b) = f(b) g(a) - f(a) g(b)$ . If $g' \neq 0$ on $(a, b)$ , Rolle’s theorem applied to $g$ itself shows $g(b) \neq g(a)$ , so the division is allowed. Taking $g(x) = x$ gives back the ordinary mean value theorem.

In plain words Think of $(g(t), f(t))$ as a point moving along a curve. Somewhere the direction of motion is parallel to the chord joining the start and the end. This version is the key step in proving l’Hôpital’s rule.

Example Take $f(x) = x^2$ and $g(x) = x^3$ on $[0, 1]$ . The equation is

(1 - 0) \cdot 3c^2 = (1 - 0) \cdot 2c,

so $c(3c - 2) = 0$ , and the solution in $(0, 1)$ is $c = \frac{2}{3}$ .

Constant functions

Definition If $f'(x) = 0$ for every $x$ in an open interval $I$ , then $f$ is constant on $I$ . To prove it, take any $x_1 < x_2$ in $I$ . The function is differentiable, hence continuous, on $[x_1, x_2]$ , so the mean value theorem gives a $c$ between them with

f(x_2) - f(x_1) = f'(c)(x_2 - x_1) = 0.

The same holds on any interval if $f$ is continuous on it and $f' = 0$ at its interior points.

In plain words A function whose slope is zero everywhere on an interval cannot change. The word interval matters. On a domain with a gap, the function can take different constant values on the separate pieces.

Example The function $f(x) = \arctan x + \arctan\left(\frac{1}{x}\right)$ has

f'(x) = \frac{1}{1 + x^2} - \frac{1}{x^2} \cdot \frac{1}{1 + 1/x^2} = \frac{1}{1 + x^2} - \frac{1}{x^2 + 1} = 0

for every $x \neq 0$ . It is constant on $(0, \infty)$ , where $f(1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$ , and constant on $(-\infty, 0)$ , where $f(-1) = -\frac{\pi}{2}$ . It is not constant on its whole domain.

Corollaries

Definition Let $I$ be an open interval.

If $f'(x) = g'(x)$ for all $x$ in $I$ , then $f - g$ is constant on $I$ , so $f(x) = g(x) + C$ for some constant $C$ .
If $f$ is differentiable on $I$ and $\lvert f'(x) \rvert \le M$ for all $x$ in $I$ , then $\lvert f(x_2) - f(x_1) \rvert \le M \lvert x_2 - x_1 \rvert$ for all $x_1, x_2$ in $I$ .
If $f$ is differentiable on $I$ and $f'(x) \ge m$ for all $x$ in $I$ , then $f(x_2) - f(x_1) \ge m(x_2 - x_1)$ whenever $x_1 < x_2$ in $I$ .

Each follows by applying the mean value theorem on $[x_1, x_2]$ .

In plain words A bound on the derivative gives a bound on how much the function can change. Two functions with the same derivative differ by a constant, which is why antiderivatives come with a $+ C$ .

Example Suppose $f$ is differentiable on an open interval containing $[1, 4]$ , with $f(1) = 10$ and $f'(x) \ge 2$ there. Then $f(4) - f(1) \ge 2 \cdot 3$ , so $f(4) \ge 16$ . Also, since $\lvert \cos x \rvert \le 1$ , the second corollary with $I = \mathbb{R}$ gives

\lvert \sin x_2 - \sin x_1 \rvert \le \lvert x_2 - x_1 \rvert \qquad \text{for all real } x_1, x_2.

Check yourself

Find the value $c$ promised by the mean value theorem for $f(x) = x^2 + 2x$ on $[0, 3]$ .

Show the answer

Since $f(0) = 0$ and $f(3) = 15$ , the average rate is

\frac{15 - 0}{3 - 0} = 5.

The derivative is $f'(x) = 2x + 2$ , and solving $2c + 2 = 5$ gives $c = \frac{3}{2}$ , which is in $(0, 3)$ .

Does Rolle’s theorem apply to $f(x) = \frac{1}{x^2}$ on $[-1, 1]$ ?

Show the answer

No. Although $f(-1) = f(1) = 1$ , the function is not defined at $0$ , so it is not continuous on $[-1, 1]$ . In fact $f'(x) = -\frac{2}{x^3}$ is never $0$ .

Show that $x^3 + x - 1 = 0$ has exactly one real solution.

Show the answer

Let $f(x) = x^3 + x - 1$ . Since $f(0) = -1 < 0 < 1 = f(1)$ , the intermediate value theorem gives at least one root. If there were two roots $r < s$ , Rolle’s theorem would give $c$ in $(r, s)$ with $f'(c) = 0$ , but $f'(x) = 3x^2 + 1 > 0$ for every $x$ . So there is exactly one.

The Mean Value Theorem

Theory, part 69

Practice, part 70

Key terms

Rolle’s theorem

Mean value theorem

The hypotheses

Cauchy’s mean value theorem

Constant functions

Corollaries

Check yourself

How this lesson connects

Builds on