The mean value theorem says that a smooth enough function must, at some moment, change at exactly its average rate. If $f$ 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 $a < c < b$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}.$$ The right side is the slope of the secant line joining the two endpoints of the graph. The left side is the slope of the tangent line at $c$. So somewhere between $a$ and $b$ the tangent line is parallel to the secant line.
Without the fog
Think of a drive. OpenStax Calculus Volume 1 uses this picture: if you drive for one hour along a straight road and your average velocity is 45 miles per hour, then, assuming your position changes smoothly, at some moment your speedometer read exactly 45. You cannot average 45 by always going slower than 45, and you cannot jump from 40 to 50 without passing through 45 if your speed changes smoothly.
The theorem is an existence statement. It does not say when that moment happened, and there can be several such moments. It only says at least one exists.
Rolle's theorem: the flat case
OpenStax presents Rolle's theorem first. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there is at least one $c$ in $(a,b)$ with $f'(c) = 0$. A smooth graph that starts and ends at the same height must turn around somewhere, and at the turning point the tangent is flat.
The mean value theorem is Rolle's theorem tilted. Subtract the secant line from $f$. The new function has equal values at both endpoints, so Rolle's theorem gives a point where its derivative is zero, which is exactly a point where $f'(c)$ equals the secant slope. That is the proof OpenStax gives.
Worked example 1: a parabola
Take $f(x) = x^2$ on $[1,3]$. It is a polynomial, so both conditions hold. The secant slope is $$\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4.$$ The derivative is $f'(x) = 2x$, so set $2c = 4$ and get $c = 2$, which is inside $(1,3)$. At $x = 2$ the tangent line has slope 4, the same as the secant line from $(1,1)$ to $(3,9)$.
Worked example 2: a cubic
Take $f(x) = x^3$ on $[0,2]$. The secant slope is $\frac{8 - 0}{2 - 0} = 4$. The derivative is $f'(x) = 3x^2$, so $3c^2 = 4$ gives $c = \pm \frac{2}{\sqrt{3}}$. Only the positive value lies in $(0,2)$, so $c = \frac{2}{\sqrt{3}} \approx 1.155$. The theorem promised at least one point, and here there is exactly one in the interval.
Worked example 3: when a condition fails
Take $f(x) = |x|$ on $[-1,1]$. It is continuous, and $f(-1) = f(1) = 1$, so the average slope is 0. But $f'(x) = -1$ for $x < 0$ and $f'(x) = 1$ for $x > 0$, and there is no derivative at $x = 0$. No point has slope 0. OpenStax makes the same point with $|x| - 1$ for Rolle's theorem: one corner is enough to break the conclusion. Continuity matters too. A function with a jump can have a large average slope while every slope you can measure is zero, as with a step function.
The speeding ticket problem
OpenStax ends the section with a classic exercise. You pass one police car at 10:17 a.m. and a second one 39 miles further on at 10:53 a.m., both times at 55 miles per hour, on a road with a 60 mile per hour limit. The trip took 36 minutes, which is 0.6 hours, so your average speed was $39 / 0.6 = 65$ miles per hour. If your position is a differentiable function of time, the mean value theorem says that at some moment your speed was exactly 65. You were over the limit at least once, even though both officers saw you at 55.
Three consequences you will use
OpenStax lists three corollaries that the rest of calculus leans on.
- Zero derivative means constant. If $f'(x) = 0$ for every $x$ in an interval, then $f$ is constant there.
- Same derivative means a constant difference. If $f'(x) = g'(x)$ on an interval, then $f(x) = g(x) + C$ for some constant $C$. This is why every antiderivative carries a $+C$.
- Sign of the derivative gives direction. If $f'(x) > 0$ on $(a,b)$, $f$ is increasing on $[a,b]$; if $f'(x) < 0$, it is decreasing. This is the basis of the first derivative test.
A version of the theorem with two functions, often called the Cauchy mean value theorem, is the usual route to proving L'Hôpital's rule.
Checkpoints
- Find every $c$ promised by the mean value theorem for $f(x) = x^2 + 2x$ on $[0,2]$.
- Does the theorem apply to $f(x) = 1/x$ on $[-1,1]$? Explain which condition fails.
- A ball is dropped and its height is $s(t) = 100 - 16t^2$ feet. It lands at $t = 2.5$ seconds. When does its velocity equal its average velocity?
- Two runners start and finish a race at the same times. Use Rolle's theorem on the difference of their positions to show that at some moment they had the same speed.
Common mistakes
- Skipping the conditions. Check continuity on the closed interval and differentiability on the open interval before you solve for $c$.
- Keeping a $c$ outside the interval. Solving $f'(c) = $ slope can give solutions outside $(a,b)$. The theorem only speaks about points strictly inside.
- Reading the theorem as "the derivative equals the average everywhere". It promises one point, not every point.
- Thinking the converse holds. A function can have a tangent parallel to the secant without meeting the conditions. Failing a condition means the guarantee is gone, not that the conclusion must fail.
Related reading
- The intermediate value theorem, an existence result about function values rather than derivatives
- The fundamental theorem of calculus, which the mean value theorem helps prove
- What a derivative actually measures
- L'Hôpital's rule intuition and epsilon delta intuition
- Calculus foundations topic map and the AP Calculus AB bridge
- Glossary: derivative, continuity, all explainers and the sources page
Related tools
The numerical derivative calculator estimates $f'(c)$ so you can check a value you found by hand. Compute the secant slope yourself first, then compare.
Sources
Last reviewed: 12 November 2026.
Common questions
What does the mean value theorem say in simple terms?
If a function is continuous on a closed interval and differentiable inside it, then somewhere inside the interval the instantaneous rate of change equals the average rate of change over the whole interval. On a graph, some tangent line is parallel to the secant line joining the endpoints.
Why does the mean value theorem need differentiability?
Without a derivative at every interior point, the graph can have a corner where the slope jumps past the average value without ever equaling it. The function |x| on the interval from minus 1 to 1 has average slope 0 but slopes of only minus 1 and 1.
What is the difference between Rolle's theorem and the mean value theorem?
Rolle's theorem is the special case where the function has equal values at the two endpoints, so the average slope is zero and the theorem guarantees a point where the derivative is zero. The mean value theorem allows any endpoint values.
Does the mean value theorem tell you where the point c is?
No. It guarantees that at least one such point exists inside the interval. Finding it means solving the derivative equation, and there can be more than one solution.