The Mean Value Theorem

In this lesson you’ll learn Rolle’s theorem and the mean value theorem, see how they follow from the extreme value theorem, and prove the facts calculus took for granted: that a positive derivative means increasing, and that a zero derivative means constant.

If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there is $c \in (a,b)$ with $f'(c)=0$.

Proof. By the extreme value theorem, $f$ attains a maximum $M$ and a minimum $m$ on $[a,b]$.

If $M = m$ then $f$ is constant and $f'(c)=0$ everywhere in $(a,b)$.

Otherwise at least one of $M, m$ differs from the shared endpoint value $f(a)=f(b)$, so it is attained at some interior point $c$. By the interior extremum theorem, $f'(c)=0$. ∎

Two theorems, both from earlier lessons, combined in four lines. EVT produces the point; the extremum theorem evaluates the derivative there.

Note the asymmetry in the hypotheses: continuity on the closed interval, differentiability only on the open one. That is deliberate, because differentiability at the endpoints is never needed and demanding it would exclude useful cases like $\sqrt{1-x^2}$ on $[-1,1]$.

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is $c \in (a,b)$ with

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

Some tangent is parallel to the secant. Equivalently, the instantaneous rate equals the average rate somewhere.

Proof. Let $L$ be the secant line through the endpoints and set $g = f - L$. Then $g$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $g(a)=g(b)=0$ since $L$ agrees with $f$ at both ends.

By Rolle, there is $c$ with $g'(c)=0$, that is $f'(c) = L'(c) = \frac{f(b)-f(a)}{b-a}$. ∎

So Rolle is the special case $f(a)=f(b)$, and the MVT is Rolle after tilting the picture. Subtracting off the secant to reduce to a flat case is a move worth remembering; it recurs in Taylor’s theorem next lesson.

Continuity on the closed interval. Let $f(x)=x$ on $[0,1)$ and $f(1)=0$. Then $f(0)=f(1)=0$ but $f'=1$ everywhere it exists. Rolle fails, because $f$ is not continuous at 1.

Differentiability on the open interval. Let $f(x)=|x|$ on $[-1,1]$. Then $f(-1)=f(1)=1$, but $f'=\pm1$ wherever it exists and is never 0. Rolle fails, because $f$ is not differentiable at 0.

Each counterexample removes exactly one hypothesis, which is how you know neither is decoration.

This is where the lesson pays for itself. Every one of these was used freely in calculus and none was proved there.

Zero derivative means constant. If $f'(x)=0$ for all $x \in (a,b)$ then $f$ is constant on $[a,b]$.

Proof. Take any $x_1<x_2$ in the interval. By MVT there is $c$ between them with

$$f(x_2)-f(x_1) = f'(c)(x_2-x_1) = 0$$

so $f(x_2)=f(x_1)$. All values agree. ∎

This is what justifies “+C” in integration. Two antiderivatives of the same function differ by a constant precisely because their difference has zero derivative. Without the MVT, the constant of integration is an unproved assumption.

Positive derivative means increasing. If $f'>0$ on $(a,b)$ then $f$ is strictly increasing there.

Proof. For $x_1<x_2$, MVT gives $f(x_2)-f(x_1) = f'(c)(x_2-x_1) > 0$. ∎

Also used constantly and never proved in a first course. Note the converse is weaker: a strictly increasing differentiable function has $f' \ge 0$, not $f' > 0$, as $x^3$ at 0 shows.

Bounded derivative gives Lipschitz. If $|f'|\le K$ on an interval then $|f(x)-f(y)|\le K|x-y|$.

That is the result quoted in the uniform continuity lesson, and the MVT is its proof: $|f(x)-f(y)| = |f'(c)||x-y| \le K|x-y|$.

If $f,g$ are continuous on $[a,b]$ and differentiable on $(a,b)$, there is $c$ with

$$\big(f(b)-f(a)\big)g'(c) = \big(g(b)-g(a)\big)f'(c)$$

Taking $g(x)=x$ recovers the ordinary MVT. This generalized version is what proves L’Hôpital’s rule, another calculus staple that gets its justification here rather than where it was taught.

Example 1: Verify the MVT for $f(x)=x^2$ on $[1,3]$.

Solution. Continuous and differentiable everywhere, so MVT applies.

$$\frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = 4$$

Solve $f'(c)=2c=4$, giving $c=2$, which lies in $(1,3)$. ✓

For any quadratic the point $c$ is the midpoint of the interval, which is a pleasant special fact and not true in general.

Example 2: Show Rolle fails without differentiability.

Solution. $f(x)=|x|$ on $[-1,1]$: continuous on $[-1,1]$, and $f(-1)=f(1)=1$.

But $f'(x)=-1$ for $x<0$ and $+1$ for $x>0$, and $f'(0)$ does not exist. So there is no $c$ with $f'(c)=0$.

Rolle fails, and it fails only because of the single non-differentiable point.

Example 3: Prove $\sin$ is Lipschitz with constant 1.

Solution. For any $x<y$, MVT gives $c \in (x,y)$ with

$$\sin y - \sin x = \cos(c)(y-x)$$

Since $|\cos c|\le1$,

$$|\sin y - \sin x| \le |y-x|$$

∎

This is the estimate used in the uniform continuity lesson, now proved. It also gives the familiar bound $|\sin x| \le |x|$ by taking $y=x$, $x=0$.

Example 4: Use the MVT to bound a value.

Given $f(1)=5$ and $|f'(x)|\le 3$ for all $x$, bound $f(4)$.

Solution. MVT on $[1,4]$ gives $c$ with

$$f(4)-f(1) = f'(c)(4-1) = 3f'(c)$$

Since $|f'(c)|\le3$, we get $|f(4)-5|\le9$, so

$$-4 \le f(4) \le 14$$

∎

This is the standard way a derivative bound converts into a value bound, and it is the mechanism behind numerical error estimates.

Example 5: Prove an inequality.

Show $e^x \ge 1+x$ for all $x$.

Solution. Let $g(x)=e^x-1-x$, so $g(0)=0$ and $g'(x)=e^x-1$.

For $x>0$: $g'>0$ on $(0,x)$, so by the MVT consequence $g$ is increasing, giving $g(x)>g(0)=0$.

For $x<0$: $g'<0$ on $(x,0)$, so $g$ is decreasing there, meaning $g(x)>g(0)=0$.

Either way $g(x)\ge0$, that is $e^x\ge1+x$, with equality only at $x=0$. ∎

Monotonicity from the derivative’s sign is the standard tool for proving inequalities, and the MVT is what licenses it.

Example 6: At most one root.

Show $x^3+x-1=0$ has exactly one real root.

Solution. At least one. By IVT, since $f(0)=-1<0$ and $f(1)=1>0$.

At most one. $f'(x)=3x^2+1 > 0$ everywhere, so $f$ is strictly increasing and cannot take the value 0 twice.

Exactly one root. ∎

Alternatively via Rolle: two roots would force $f'(c)=0$ somewhere between them, and $f'$ is never 0.

Example 7: Why “+C” is a theorem.

If $F'=G'$ on an interval, show $F-G$ is constant.

Solution. Let $H = F-G$. Then $H' = F'-G' = 0$ on the interval.

By the MVT consequence, $H$ is constant, so $F = G + C$. ∎

Every antiderivative of a given function differs from every other by a constant, and this is why. Note it needs the domain to be an interval: on $(-\infty,0)\cup(0,\infty)$ a function can have zero derivative and take different constants on each piece, which is why $\int\frac{1}{x}dx$ carries different constants on either side of 0.

Speed cameras. Average-speed enforcement over a stretch of road relies on the MVT: if your average speed exceeded the limit, then at some instant your actual speed did too. That is a legally meaningful application of the theorem.

Numerical error bounds. Taylor and finite-difference error terms are stated with an unspecified interior point, and they come directly from MVT-style arguments. Bounding the derivative there gives the usable error estimate.

Root-finding uniqueness. Newton’s method converges to a root; showing there is only one, so you know which, is a monotonicity argument via the MVT. Example 6 is the pattern.

Physics and conservation. Proving a quantity is conserved amounts to showing its derivative is zero and concluding it is constant. That step is a MVT consequence, not a definition.

Economics. The claim that a marginal quantity somewhere equals the average over an interval is used in pricing and in proving existence of representative agents. It is the MVT with an economic label.

Signal analysis. A bound on a signal’s rate of change limits how much it can move between samples, which is what determines a safe sampling rate. Converting a derivative bound into a value bound is Example 4.