Mean Value Theorem¶

Theorem¶

Let $a < b$ be real numbers, and consider a function $f\colon [a, b] \to \mathbb{R}$ that is continuous on $[a, b]$ and differentiable on $(a, b)\text{.}$ Then, we can assert that there exists a $c \in (a, b)$ such that

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

Interpretation¶

The mean value theorem links the average rate of change of $f$ over an interval with its instantaneous rate of change: somewhere strictly between $a$ and $b$ , the derivative equals the slope of the secant through $(a,f(a))$ and $(b,f(b))$ . If a car covers $100$ km in one hour, then at some moment its speed is exactly $100$ km/h.

The cubic x cubed minus x on an interval from a to b with the secant through its endpoints and two tangents at points c1 and c2 that are parallel to the secant.

The figure shows $f(x)=x^3-x$ on $[-2,2]$ . The secant has slope $(f(2)-f(-2))/4=(6+6)/4=3$ $3$ , and $f'(c)=3c^2-1=3$ gives $c=\pm2/\sqrt{3}\approx\pm1.15$ $\pm 1.15$ . These are the points $c_1$ and $c_2$ , where the tangents are parallel to the secant. The theorem promises at least one such point; it says nothing about how many there are or where they lie.

Move $a$ and $b$ : for a preset satisfying the theorem’s hypotheses there is a point $c\in(a,b)$ where the tangent is parallel to the secant. With $|x|$ and $0\in(a,b)$ the hypothesis fails, and so can the conclusion.

Both hypotheses are needed. Without continuity at an endpoint the conclusion can fail: $f(x)=x$ on $[0,1)$ with $f(1)=0$ has secant slope $0$ , but $f'(x)=1$ on $(0,1)$ . Without differentiability at a single interior point it can fail too: for $|x|$ on $[-1,2]$ the secant slope is $\tfrac{1}{3}$ , while the derivative only takes the values $-1$ and $1$ . Differentiability at the endpoints themselves is not required.

Rolle’s theorem is the special case $f(a)=f(b)$ $f (b)$ , and the general theorem follows from it by subtracting the secant line from $f$ . In practice the theorem is used in the form $f(y)-f(x)=f'(c)(y-x)$ from the Key Results, which turns information about $f'$ into information about $f$ : the sign of $f'$ gives monotonicity, and a bound $|f'|\le M$ on an interval gives $|f(y)-f(x)|\le M|y-x|$ for all $x,y$ in it.

Key Results¶

The theorem requires continuity on a closed interval and differentiability on the open interval.
It justifies the derivative sign test. Applying the theorem on $[x,y]$ gives $f(y)-f(x)=f'(c)(y-x)$ for some $c\in(x,y)\text{,}$ so the sign of $f'$ on the interior determines whether $f$ is increasing, decreasing, or constant.
Two functions with the same derivative. If $f,g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ with $f'=g'$ on $(a,b)\text{,}$ then $f=g+C$ on $[a,b]$ for some constant $C\text{.}$ The domain has to be an interval for this to hold.
The version above is Lagrange's mean value theorem, and it is what "the mean value theorem" means in this course. The two-function version due to Cauchy, which underlies L'Hôpital's rule, is optional and non-assessable.
It generalizes Rolle’s theorem.
The conclusion does not specify the location of the point.

Examples¶

For $f(x)=x^2$ on $[0,1]$ , the secant slope is $(1-0)/(1-0)=1$ $1$ . Solving $f'(c)=2c=1$ gives $c=1/2$ $1/2$ .
For $f(x)=3x+2$ on any interval $[a,b]$ with $a<b$ $b$ , the secant slope and derivative both equal $3$ , so every $c\in(a,b)$ satisfies the conclusion.

Using the mean value theorem to prove an inequality

Prove that $|\sin a-\sin b|\le|a-b|$ for all real numbers $a$ and $b$ .

Set up an interval

If $a=b$ both sides are $0$ . Otherwise, swapping the names if needed, assume $a<b$ $b$ ; the inequality is symmetric in $a$ and $b$ . The function $f(x)=\sin x$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , with $f'(x)=\cos x$ $cos x$ .

Apply the theorem

The mean value theorem gives a $c\in(a,b)$ with

\cos c=\frac{\sin b-\sin a}{b-a}.

Bound the derivative

Since $|\cos c|\le1$ $1$ ,

|\sin b-\sin a|=|\cos c|\,|b-a|\le|b-a|.

Takeaway. The mean value theorem turns a bound on $f'$ into a bound on differences of $f$ : if $|f'|\le M$ on an interval, then $|f(b)-f(a)|\le M|b-a|$ there. You never need to know where $c$ is.

Prerequisites¶

Rolle’s Theorem
Monotone Function: the sign of the derivative determines the sign of each secant slope.

Sources¶

Math I: Calculus · Lecture 7

Aliases: MVT, Lagrange's mean value theorem, Lagrange mean value theorem, law of the mean Symbols: element of, real numbers, square root

Last updated: 27 October 2026 11:52