Mean Value Theorem: Interactive Visualization

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Mean value theorem in this visualization

The mean value theorem, also called Lagrange’s mean value theorem, says: if a function $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ , then there is at least one point $c$ with $a < c < b$ where

Mean value theorem

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

The left side is the slope of the tangent at $c$ ; the right side is the slope of the secant through $(a, f(a))$ and $(b, f(b))$ , the average rate of change of $f$ over the interval. The point is also written $\xi$ . Rolle’s theorem is the case $f(a) = f(b)$ , where the secant is flat and $f'(c) = 0$ . The page shows the theorem on four smooth functions and on two whose hypotheses can fail: $|x|$ is not differentiable at its corner, and the jump function, on an interval with the jump inside, is neither continuous nor differentiable there.

The blue points A and B sit on the curve above $a$ and $b$ : drag them, or slide the blue segment on the x-axis to move the whole interval. The yellow line through them is the secant; call its slope $m$ . Each blue point marks a point $c$ : its tangent is parallel to the secant, and a dashed vertical joins it to the secant: where $f$ is smooth, the height of the curve above the secant has derivative zero at every $c$ (the two have the same slope there). Under the graph, the slope strip draws $f'(x)$ in blue and the secant slope $m$ as a yellow horizontal line: every $c$ is a crossing. With the answers hidden, a tangent point appears that you can move along the curve to find $c$ yourself.

02

What not to read into it

The theorem promises at least one $c$ , not exactly one: $x^3 - 3x$ on $[-2, 2]$ has two, $\sin x$ on $[-6, 6]$ has four. It does not say where $c$ is: for a non-degenerate quadratic the unique $c$ is always the midpoint, but for other functions it can lie anywhere inside.
Differentiability is needed only inside the interval. $\sqrt{x}$ on $[0, 4]$ has a vertical tangent at 0, and still $c = 1$ works.
When a hypothesis fails, the theorem is silent. $|x|$ on $[-1, 2]$ has no $c$ , nor does the jump function on $[0.5, 1.5]$ ; yet the same jump function on $[0, 2]$ happens to have $c = 1.5$ .
The points $c$ on this page are solved exactly from $f'(c) =$ secant slope for each function, not found by a numerical search.

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Where it is used

Average speed: a car covers 10 km in 5 minutes, an average of 120 km/h. If the distance it has travelled changes continuously and has a derivative, the speed, at every moment inside those 5 minutes, then at some instant its speed was exactly 120 km/h. So a driver whose average over a camera-timed stretch is above the limit was over the limit at some moment.
If $f' = 0$ everywhere on an interval, $f$ is constant there; two functions with the same derivative differ by a constant, which is why an antiderivative carries $+\,C$ .
Estimates: if $|f'| \le M$ on the interval, then $|f(b) - f(a)| \le M\,|b - a|$ . The theorem is also a step in the proof of the fundamental theorem of calculus.

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Mean value theorem in this visualization

What not to read into it

Where it is used

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