Prove the Mean Value Theorem

From Equal Endpoint Values to a Secant Slope

Rolle’s Theorem guarantees an interior point where the derivative is zero when the endpoint values agree. The Mean Value Theorem removes that equality condition: for a continuous function on a closed interval, differentiable in its interior, some interior derivative equals the slope of the line joining the endpoint values. The key proof idea is to subtract that line from the function. The difference then has equal endpoint values, so Rolle’s Theorem applies.

The hypotheses are the same as in Rolle’s Theorem: continuity on the entire closed interval and differentiability at every interior point. The conclusion is different. Rather than finding a point with derivative zero, we find a point where the instantaneous rate of change matches the average rate of change over the interval.

Theorem (Mean Value Theorem): Let $a<b$, and let $f:[a,b]\to\mathbb R$ be continuous on $[a,b]$ and differentiable at every point of $(a,b)$. Then there is a $c\in(a,b)$ such that $$ f'(c)=\frac{f(b)-f(a)}{b-a}. $$

The quotient on the right is the slope of the secant line through $(a,f(a))$ and $(b,f(b))$. The theorem says that the graph has an interior point where its tangent has that same slope. It does not claim that this point is unique.

Proof Using an Auxiliary Function

Let $m=\frac{f(b)-f(a)}{b-a}$, the secant slope. Consider the line through the two endpoint values, written as $L(x)=f(a)+m(x-a)$. Subtract this line from $f$ and define $g(x)=f(x)-L(x)$. At $a$, the line has value $f(a)$, so $g(a)=0$. At $b$, the definition of $m$ gives $L(b)=f(a)+m(b-a)=f(b)$, so $g(b)=0$ as well.

Because $f$ and $L$ are continuous on $[a,b]$, their difference $g$ is continuous there. Because $f$ and $L$ are differentiable on $(a,b)$, $g$ is differentiable there. Thus $g$ satisfies every hypothesis of Rolle’s Theorem.

Proof of the Mean Value Theorem. By Rolle’s Theorem, there is a $c\in(a,b)$ such that $g'(c)=0$. Since $g(x)=f(x)-f(a)-m(x-a)$, differentiation at $c$ gives $g'(c)=f'(c)-m$. Therefore $0=f'(c)-m$, so

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

This is the required conclusion. $\square$

The auxiliary function is sometimes described as removing the secant slope: its derivative is $f'(x)-m$, so a zero of that derivative is exactly a point where $f'(x)=m$. The endpoint calculation is essential. It is what makes Rolle’s Theorem applicable to the modified function.

Worked Examples

Worked Example: A Quadratic on an Asymmetric Interval

Let $f(x)=x^2$ on $[1,4]$. A polynomial is continuous on the closed interval and differentiable at every interior point, so the Mean Value Theorem applies. The endpoint values are $f(1)=1^2=1$ and $f(4)=4^2=16$. Hence the secant slope is

\frac{f(4)-f(1)}{4-1}=\frac{16-1}{3}=5.

Since $f'(x)=2x$, the theorem’s conclusion requires $2c=5$, giving $c=5/2$. This point is in the open interval because $1<5/2<4$. Direct substitution verifies the conclusion: $f'(5/2)=2(5/2)=5$, which equals the secant slope.

Worked Example: A Logarithm on a Long Interval

Take $f(x)=\ln x$ on $[1,e^2]$. This function is continuous on that closed interval and differentiable in its interior. Its endpoint values are $\ln 1=0$ and $\ln(e^2)=2$, so the secant slope is

\frac{\ln(e^2)-\ln 1}{e^2-1}=\frac{2}{e^2-1}.

The derivative is $f'(x)=1/x$. Equating it to the secant slope gives $1/c=2/(e^2-1)$, and therefore $c=(e^2-1)/2$. This point belongs to the required interval: since $e^2>3$, we have $(e^2-1)/2>1$; and since $e^2>0$, we have $e^2-1<2e^2$, so $(e^2-1)/2<e^2$. Substitution confirms $f'(c)=1/c=2/(e^2-1)$, exactly the secant slope.

Worked Example: A Trigonometric Function

Let $f(x)=\sin x$ on $[0,\pi/2]$. It is continuous on the closed interval and differentiable in the interior. The endpoint values are $\sin 0=0$ and $\sin(\pi/2)=1$, so the secant slope is

\frac{1-0}{\pi/2-0}=\frac{2}{\pi}.

Because $f'(x)=\cos x$, the point supplied by the theorem satisfies $\cos c=2/\pi$. In this case, $c=\arccos(2/\pi)$. This is an interior point: $0<2/\pi<1$, since $\pi>2$, and the cosine decreases from $1$ to $0$ on $[0,\pi/2]$. Thus $0<c<\pi/2$. Substitution gives $f'(c)=\cos c=2/\pi$, as required.

A Derivative Bound Controls Endpoint Change

The Mean Value Theorem gives more than an existence statement about a matching slope. If every interior derivative is bounded in absolute value, then the change in the function between the endpoints cannot exceed that bound multiplied by the interval length.

Theorem (Endpoint-Change Bound): Let $a<b$, and let $f:[a,b]\to\mathbb R$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose $|f'(x)|\leq M$ for every $x\in(a,b)$, where $M\geq0$. Then $$ |f(b)-f(a)|\leq M(b-a). $$

Proof. By the Mean Value Theorem, there is a $c\in(a,b)$ such that $f(b)-f(a)=f'(c)(b-a)$. Taking absolute values and using $b-a>0$ gives

|f(b)-f(a)|=|f'(c)|(b-a)\leq M(b-a).

This proves the bound. $\square$

The same result applies to any two points $x<y$ in an interval on which the hypotheses hold: apply the theorem to the restricted interval $[x,y]$. Thus a uniform bound on the derivative controls differences in function values throughout the interval, not just across one originally chosen pair of endpoints.

Worked Example: Bounding the Change in the Square Root

Consider $f(x)=\sqrt{x}$ on $[4,9]$. Its derivative is $f'(x)=1/(2\sqrt{x})$, and throughout the open interval $4<x<9$ we have $0<f'(x)<1/4$, because $\sqrt{x}>2$. In particular, $|f'(x)|\leq1/4$. The Endpoint-Change Bound therefore gives

|\sqrt{9}-\sqrt{4}|\leq\frac14(9-4)=\frac54.

The left side is $|3-2|=1$, and $1\leq5/4$, so the claimed estimate checks out. The bound is valid but not sharp here; the theorem does not promise the smallest possible value of $M$ or the tightest estimate.

Why Both Endpoint Continuity and Interior Differentiability Matter

The Mean Value Theorem’s assumptions have separate jobs. Continuity on the closed interval allows the use of Rolle’s Theorem on the auxiliary function, while differentiability in the interior allows its derivative to be calculated and forces the matching-slope conclusion. A function may be differentiable at every interior point yet still fail to be continuous at an endpoint. That failure can invalidate the conclusion.

For example, define $f:[0,1]\to\mathbb R$ by $f(0)=2$ and $f(x)=x$ for $x\neq0$. The point $0$ is an endpoint, so the discontinuity there does not violate the interior differentiability hypothesis. At every $x\in(0,1)$, the function agrees with $x$ in a neighborhood of $x$, and therefore $f'(x)=1$. But the secant slope across the full interval is

\frac{f(1)-f(0)}{1-0}=\frac{1-2}{1}=-1.

There is no interior point where $f'(c)=-1$, because its derivative is $1$ throughout $(0,1)$. The function fails continuity on $[0,1]$, precisely at the endpoint $0$, so this example does not contradict the Mean Value Theorem.

Conversely, endpoint continuity alone is not enough: differentiability at every interior point is also required. For $f(x)=|x|$ on $[-1,1]$, the secant slope is zero, but the derivative is $-1$ for $-1<x<0$ and $1$ for $0<x<1$, and it does not exist at $0$. No differentiable interior point has derivative zero. The missing differentiability at $0$ prevents the theorem from applying.

A common proof error is to use the endpoint quotient without first checking the hypotheses, or to claim that the theorem identifies a particular point without solving the equation $f'(c)=\frac{f(b)-f(a)}{b-a}$. The theorem guarantees at least one such point; it may not identify it uniquely. When a specific point is proposed, verify both that it lies strictly inside the interval and that its derivative really equals the calculated secant slope.

Compute the average slope.
Find the endpoint quotient $\frac{f(b)-f(a)}{b-a}$, keeping the order of the endpoint values and interval length consistent.

Check the hypotheses.
Verify continuity on the closed interval and differentiability at every interior point.

Match the derivative to the secant slope.
Use the theorem to conclude that an interior point exists, or solve the resulting derivative equation when a point is needed.

Check Your Understanding

Use the statement, proof, and applications of the Mean Value Theorem to answer these questions.

What auxiliary function turns the Mean Value Theorem into an application of Rolle’s Theorem, and why do its endpoint values agree?
For $f(x)=x^2$ on $[1,4]$, calculate the secant slope and verify the interior point where the derivative equals it.
Why does the function defined by $f(0)=2$ and $f(x)=x$ for $x\neq0$ on $[0,1]$ not contradict the Mean Value Theorem?
State the Endpoint-Change Bound and identify the theorem used to prove it.
Why does the Mean Value Theorem guarantee at least one matching-slope point but not necessarily a unique one?

What You'll Learn