The Mean Value Theorem

The chord

Take a smooth arc from x = a to x = b and join its two ends with a straight line, the chord. Its gradient is the rise over the run, $\left(f\left(b\right)-f\left(a\right)\right)/\left(b-a\right)$(f(b) − f(a)) / (b − a): the average rate at which f changes from a to b.

For $y=\frac{x^2}{4}$y = x²/4 from 0 to 4, the ends are (0, 0) and (4, 4), so the chord has gradient (4 − 0) ÷ (4 − 0) = 1.

A tangent parallel to the chord

Slide a line parallel to the chord across the arc. At the ends the arc may be steeper or shallower than the chord, but somewhere in between the sliding line just touches the arc, and there it is a tangent with the chord's gradient.

That is the mean value theorem: if f is continuous on [a, b] and differentiable on (a, b), then there is at least one c with a < c < b where $f^{\prime }\left(c\right)=\left(f\left(b\right)-f\left(a\right)\right)/\left(b-a\right)$f'(c) = (f(b) − f(a)) / (b − a). The instantaneous rate at c equals the average rate over the whole interval.

For $y=\frac{x^2}{4}$y = x²/4 on [0, 4], the derivative is $\frac{x}{2}$x/2, and setting $\frac{c}{2}$c/2 equal to the chord gradient 1 gives c = 2, strictly between 0 and 4. The tangent at (2, 1) is y = x − 1, parallel to the chord y = x.

The arc $y=\frac{x^2}{4}$y = x²/4 from (0, 0) to (4, 4), the chord y = x joining its ends, and the tangent y = x − 1 at c = 2. Both lines have gradient 1.

Finding c

To find c, work out the chord gradient, set f'(c) equal to it, solve, and keep only the solutions strictly between a and b.

Take $f\left(x\right)=x^3-x$f(x) = x³ − x on [0, 2]. The ends are f(0) = 0 and f(2) = 8 − 2 = 6, so the chord gradient is (6 − 0) ÷ 2 = 3. The derivative is $3x^2-1$3x² − 1, and $3c^2-1=3$3c² − 1 = 3 gives $c^2=\frac{4}{3}$c² = 4/3, so $c=\pm \frac{2}{\sqrt{3}}$c = ±2/√3. Only the positive value, about 1.154701, lies in (0, 2). Check: $3\times \frac{4}{3}-1=3$3 × 4/3 − 1 = 3.

On [−2, 2] the same function has f(−2) = −6 and f(2) = 6, so the chord gradient is 12 ÷ 4 = 3 again, and now both $c=\frac{2}{\sqrt{3}}$c = 2/√3 and $c=-\frac{2}{\sqrt{3}}$c = −2/√3 lie inside. The theorem promises at least one c; there can be more.

Take $f\left(x\right)=\sqrt{x}$f(x) = √x on [1, 9]. The chord gradient is $\left(3-1\right)÷8=\frac{1}{4}$(3 − 1) ÷ 8 = 1/4. The derivative is $\frac{1}{2\sqrt{x}}$1/(2√x), and $\frac{1}{2\sqrt{c}}=\frac{1}{4}$1/(2√c) = 1/4 gives $\sqrt{c}=2$√c = 2, so c = 4, inside (1, 9).

The curve $y=x^3-x$y = x³ − x on [−2, 2], with the chord from (−2, −6) to (2, 6), gradient 3, drawn softer. The tangents at $x=\frac{2}{\sqrt{3}}$x = 2/√3 and $x=-\frac{2}{\sqrt{3}}$x = −2/√3 both have gradient 3, parallel to the chord.

Both conditions are needed

Continuous on [a, b] means the arc is drawn without lifting the pen, ends included. Differentiable on (a, b) means it has a tangent at every point strictly between; the ends may be corners.

Drop differentiability and the theorem fails. y = |x| on [−1, 1] is continuous, and both ends are at height 1, so the chord is flat. But the gradient is −1 everywhere left of 0 and 1 everywhere right of it, and at 0 there is a corner with no tangent at all. No tangent is flat.

Drop continuity and it fails too. Let f(x) = x for $0\le x<1$0 ≤ x < 1 and f(1) = 0. The ends are both at height 0, so the chord is flat, but every tangent between has gradient 1. The jump at x = 1 is what breaks it.

The curve y = |x| on [−1, 1] and its flat chord at height 1. Every tangent has gradient −1 or 1, and the corner at the origin has none.

Rolle's theorem

When f(a) = f(b), the chord is flat and its gradient is 0, so the theorem promises a c strictly inside with f'(c) = 0: a level tangent. This special case is Rolle's theorem.

For $f\left(x\right)=x^2-1$f(x) = x² − 1 on [−1, 1], both ends are at height 0. The derivative 2x is 0 at c = 0, inside (−1, 1), where the curve has its lowest point, (0, −1), and its tangent is the level line y = −1.

What it is used for

The theorem is mostly used to reason about a function from its derivative. If f'(x) = 0 at every point of an interval, then for any two points a and b in it, f(b) − f(a) = f'(c)(b − a) = 0, so f is constant there. If f'(x) > 0 throughout, then f(b) − f(a) is positive whenever b > a, so f is increasing. That second fact is the next lesson's test for increasing and decreasing functions.

The usual mistakes

Giving the rise or the run as the rate. With f(1) = 2 and f(5) = 14, the rise is 12 and the run is 4; the chord gradient, and so f'(c), is 12 ÷ 4 = 3.

Giving the chord gradient as c. For $x^2$x² on [0, 4] the chord gradient is 4, and c is where f' reaches it: 2c = 4, so c = 2.

Taking c at an end. f'(4) = 8 for $x^2$x² on [0, 4] is the gradient at the right-hand end; the theorem's c lies strictly inside.

Reading the theorem as "the rate equals the average all the way". It names at least one instant, no more.

Two speed cameras

In the application below, two cameras 12 km apart time a van at 8 minutes, an average of 90 km/h. The theorem says that at some instant the van was doing exactly that speed, and setting the derivative of its distance equal to the average finds the minute.