If you drive 120 miles in 2 hours, at some moment your speedometer read exactly 60 mph. The Mean Value Theorem turns that common sense into mathematics. It needs Where Derivatives Fail to Exist on the interval, and it builds on the facts about extremes from Critical Points and Absolute Extrema.
Key idea
Mean Value Theorem (MVT). If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there is at least one $c$ in $(a, b)$ with $$f'(c) = \frac{f(b) - f(a)}{b - a}.$$ Somewhere the tangent line is parallel to the secant line through the endpoints.
Why it works
First a special case, Rolle's theorem: if also $f(a) = f(b)$, then some $f'(c) = 0$. A continuous $f$ on $[a, b]$ has a max and a min; if both occur at the endpoints, $f$ is constant and $f' = 0$ everywhere; otherwise one occurs inside, where Fermat's theorem gives $f'(c) = 0$. For the MVT, subtract the secant line: $g(x) = f(x) - \left[f(a) + m(x - a)\right]$ with $m = \frac{f(b) - f(a)}{b - a}$. Then $g(a) = g(b) = 0$, Rolle gives $g'(c) = 0$, and $g'(c) = f'(c) - m$.
Worked example: Find the promised $c$
Find $c$ for $f(x) = x^3$ on $[0, 3]$.
- Secant slope: $\frac{27 - 0}{3} = 9$.
- $f'(c) = 3c^2 = 9$, so $c^2 = 3$ and $c = \pm\sqrt{3}$.
- Only $\sqrt{3} \approx 1.73$ lies in $(0, 3)$.
Answer: $c = \sqrt{3}$
Worked example: A reciprocal
Find $c$ for $f(x) = \frac{1}{x}$ on $[1, 4]$.
- $f$ is continuous and differentiable on $[1, 4]$ (0 is not in it). Secant slope: $\frac{\frac{1}{4} - 1}{3} = -\frac{1}{4}$.
- $f'(c) = -\frac{1}{c^2} = -\frac{1}{4}$ gives $c = 2$.
Answer: $c = 2$
Worked example: What the theorem proves
Show that if $f'(x) = 0$ for every $x$ in an interval, then $f$ is constant there.
- Take any two points $x_1 < x_2$ in the interval. The MVT gives a $c$ between them with $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$.
- But $f'(c) = 0$, so $f(x_2) = f(x_1)$.
- Every pair of values is equal: $f$ is constant.
Answer: So two functions with the same derivative differ by a constant. This is why antiderivatives come with "+ C".
Common mistake
Applying the theorem to a function that is not differentiable on the open interval. For $f(x) = \lvert x \rvert$ on $[-1, 2]$, the secant slope is $\frac{2 - 1}{3} = \frac{1}{3}$, but $f'$ is only ever $-1$ or $1$: no such $c$ exists. The corner at 0 breaks the hypothesis. Check continuity on $[a, b]$ and differentiability on $(a, b)$ first.