4.4 The Mean Value Theorem

Big idea

The derivative measures an instantaneous rate, while a difference quotient over an interval measures an average rate. Nothing so far connects the two. This section supplies the bridge: on a well behaved interval, the average rate is achieved exactly, at some interior instant.

The picture is a secant line through the two endpoints of the graph and a parallel tangent line somewhere between them. Slide the secant line upward without tilting it until it last touches the graph; at that final contact the line is tangent, so the slopes match. That sliding argument is the proof in outline, and it explains why the conclusion is an existence statement. You learn that such a point is there; you are not handed its location.

If you have ever been told that a car averaging $70$ mph over an hour must at some moment have been going exactly $70$ mph, you have used this theorem. The average over the interval is the difference quotient; the moment is the interior point.

The reason the result matters is not the statement but its consequences. Almost every fact connecting the sign of $f'$ to the shape of $f$ is a corollary: a function whose derivative is zero everywhere is constant, two functions with the same derivative differ by a constant, and a positive derivative means the function increases. Those are the facts the rest of the chapter, and all of integration, are built on, and none of them can be proved by staring at the definition of a derivative.

Decoder

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is at least one $c$ in $(a,b)$ with $f'(c)$ equal to the average rate of change of $f$ across $[a,b]$.

Take the two hypotheses one at a time. Continuity is required on the closed interval, endpoints included, because the secant line is drawn through those endpoints. Differentiability is required only on the open interval, because the tangent line is taken strictly inside; a function may have a corner or a vertical tangent exactly at $a$ or $b$ and still obey the theorem.

The conclusion is “at least one.” There may be several such $c$, and the theorem names none of them. Any use of this result that depends on knowing $c$ is misusing it; correct uses either bound $f'(c)$ by something known, or observe that some $c$ existing is already a contradiction.

The classic mistake is applying the theorem where differentiability fails somewhere inside the interval. The absolute value function on $[-1,1]$ has average rate $0$, but its derivative is $-1$ or $1$ everywhere it exists and is never $0$. The single bad point at the origin is enough to void the conclusion.

Definitions and results

Average rate of change. Over $[a,b]$ the average rate of change of $f$ is the slope of the secant line through the endpoints,

$$ \frac{f(b) - f(a)}{b - a} $$

Rolle’s theorem. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there is some $c$ in $(a,b)$ with $f'(c) = 0$. A function that returns to the same height must level off somewhere in between.

The mean value theorem. Under the same continuity and differentiability hypotheses, but with no assumption relating $f(a)$ and $f(b)$, there is some $c$ in $(a,b)$ with

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

How the two are related. The mean value theorem follows from Rolle’s theorem applied to the difference between $f$ and the secant line, since that difference takes equal values at the two endpoints. Rolle’s theorem is the special case where the secant is horizontal.

Zero derivative means constant. If $f'(x) = 0$ for every $x$ in an interval, then $f$ is constant there. For any two points in the interval the theorem gives a $c$ with $f(y) - f(x) = f'(c)(y - x) = 0$.

Equal derivatives mean a constant difference. If $f'(x) = g'(x)$ on an interval, then $f(x) = g(x) + C$ for some constant $C$. Apply the previous statement to $f - g$. This is the fact that makes the constant of integration both necessary and sufficient.

Sign of the derivative and monotonicity. If $f' > 0$ throughout an interval then $f$ is increasing there, and if $f' < 0$ throughout then $f$ is decreasing. Again, $f(y) - f(x) = f'(c)(y-x)$ carries the sign.

Bounding a function from its derivative. If $|f'(x)| \le M$ on an interval, then $|f(y) - f(x)| \le M|y - x|$ for all $x$ and $y$ in it. A bound on speed becomes a bound on distance travelled.

Worked examples

Finding the guaranteed point

Let $f(x) = \sqrt{x}$ on $[0,4]$. Find every $c$ the mean value theorem promises.

The function is continuous on $[0,4]$ and differentiable on $(0,4)$; the vertical tangent at $x = 0$ sits at an endpoint, so it does no harm. The average rate is

$$ \frac{f(4) - f(0)}{4 - 0} = \frac{2 - 0}{4} = \frac{1}{2} $$

Set the derivative equal to it:

$$ \frac{1}{2\sqrt{c}} = \frac{1}{2} \qquad \Longrightarrow \qquad \sqrt{c} = 1 \qquad \Longrightarrow \qquad c = 1 $$

Since $1$ lies in $(0,4)$, the theorem is satisfied, with exactly one such point. Check by drawing the secant from $(0,0)$ to $(4,2)$: it has slope $1/2$, and the tangent at $(1,1)$ has slope $1/2$ as well, so the two lines are parallel.

A horizontal tangent from equal endpoint values

Let $f(x) = x^2 - 4x + 3$ on $[1,3]$.

Here $f(1) = 1 - 4 + 3 = 0$ and $f(3) = 9 - 12 + 3 = 0$, so the endpoint values agree and Rolle’s theorem applies. Solving $f'(c) = 2c - 4 = 0$ gives $c = 2$, which is in $(1,3)$. The value there is $f(2) = 4 - 8 + 3 = -1$, the vertex of the parabola, which is where a symmetric arc must level off.

When differentiability fails inside

Let $f(x) = |x|$ on $[-1, 1]$.

The average rate of change is

$$ \frac{f(1) - f(-1)}{1 - (-1)} = \frac{1 - 1}{2} = 0 $$

so the theorem would promise a point with zero derivative. But $f'(x) = -1$ for $x < 0$ and $f'(x) = 1$ for $x > 0$, and $f'(0)$ does not exist. No such point exists. There is no contradiction: the hypothesis of differentiability on the whole open interval fails at $x = 0$, so the theorem never applied.

Using the theorem as a bound

Suppose $f$ is differentiable everywhere, $f(0) = 3$, and $f'(x) \le 2$ for every $x$. How large can $f(4)$ be?

Apply the theorem on $[0,4]$: there is some $c$ in $(0,4)$ with

$$ f(4) - f(0) = f'(c)(4 - 0) = 4f'(c) \le 4(2) = 8 $$

So $f(4) \le 3 + 8 = 11$. The bound is attained by $f(x) = 2x + 3$, so no better bound is possible from this information alone. Nothing here required knowing $c$, which is the usual way this theorem is used: the unknown interior point is immediately swallowed by an inequality.

Practice

These drills run the whole statement. Some ask you to check the hypotheses on a given interval, some ask you to produce every $c$ the conclusion guarantees, and some ask you to use the corollaries to bound a function or to argue that two functions differ by a constant.

Quiz

Five items on the hypotheses, Rolle’s theorem, locating the guaranteed point, and the corollaries about constant and monotone functions.