Mean Value Theorem

Average Rate and Instantaneous Rate

Rolle’s Theorem, proved in the previous tutorial, concerns a function that returns to its starting value: under the appropriate hypotheses, its derivative must vanish somewhere in between. The Mean Value Theorem extends that idea. Instead of requiring equal endpoint values, it identifies an interior point where the derivative matches the slope of the secant line joining the endpoints.

The theorem is a powerful connection between information about a function’s values and information about its derivative. It can show that a function’s change over an interval is governed by a derivative value inside that interval. It also supplies a reliable way to compare secant slopes, control approximation errors, and study the shape of a function.

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

The right side is the average rate of change of $f$ from $a$ to $b$, or equivalently the slope of the secant line through $(a,f(a))$ and $(b,f(b))$. The left side is the slope of the tangent line at $c$. Thus the theorem guarantees at least one interior tangent whose slope equals the secant slope. It does not, in general, say where that point is or that there is only one such point.

Finding the Point in Specific Examples

Worked Example: A Quadratic on a Closed Interval

Let $f(x)=x^2+1$ on $[1,3]$. Since $f$ is a polynomial, it is continuous on the closed interval and differentiable in its interior. The average rate of change is

\frac{f(3)-f(1)}{3-1} = \frac{(9+1)-(1+1)}{2} = \frac{8}{2} = 4.

The derivative is $f'(x)=2x$, so the point given by the theorem must satisfy

2c=4, \qquad c=2.

Indeed, $2\in(1,3)$, and $f'(2)=4$. For this function and interval, the derivative is strictly increasing, so this is the only point with the required derivative value. The Mean Value Theorem itself guarantees existence, not uniqueness.

Worked Example: An Interval with More Than One Possible Point

Take $g(x)=x^3$ on $[-2,1]$. The average rate of change is

\frac{g(1)-g(-2)}{1-(-2)} = \frac{1-(-8)}{3} = 3.

Since $g'(x)=3x^2$, the equation $g'(c)=3$ becomes

3c^2=3, \qquad c^2=1, \qquad c=-1\ \text{or}\ c=1.

Only $c=-1$ belongs to the open interval $(-2,1)$; $c=1$ is an endpoint and is not allowed by the theorem’s conclusion. Thus $c=-1$ is the interior point where the tangent slope equals the secant slope. This example also illustrates why checking that the candidate lies strictly between the endpoints matters.

The theorem does not assert that a curve is a straight line, nor that its derivative equals the average rate at every point. It asserts a specific existence claim: at least one interior point has exactly that derivative value. The hypotheses matter because the argument uses the behavior of the function across the entire closed interval, while the conclusion concerns a derivative at an interior point.

A Quantitative Estimate for Linear Approximation

The Mean Value Theorem also turns a bound on the derivative into a bound on how far a function’s increment can differ from a proposed linear increment. The next result is useful when $L$ is chosen as an approximate derivative on an interval. It bounds the error directly, without requiring a formula for the point $c$ supplied by the theorem.

Theorem (Error Bound from a Derivative Approximation): Let $a<b$, and let $f:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose $L\in\mathbb{R}$ and $\varepsilon\geq0$ satisfy $|f'(t)-L|\leq\varepsilon$ for every $t\in(a,b)$. Then $$ \left|f(b)-f(a)-L(b-a)\right|\leq\varepsilon(b-a). $$

Proof. By the Mean Value Theorem, there is $c\in(a,b)$ such that

$$ f(b)-f(a)=f'(c)(b-a). $$

Subtract $L(b-a)$ from both sides and factor out $b-a$. Since $b-a>0$, the assumed derivative bound at $c$ gives

\begin{aligned} \left|f(b)-f(a)-L(b-a)\right| &=\left|(f'(c)-L)(b-a)\right|\\ &=|f'(c)-L|(b-a)\\ &\leq\varepsilon(b-a). \end{aligned}

This proves the claimed estimate. $\square$

In this estimate, $L(b-a)$ is the change predicted by a line of slope $L$, while $f(b)-f(a)$ is the actual change. If the derivative remains close to $L$, the actual change cannot differ greatly from that linear prediction.

Worked Example: Bounding the Error in a Linear Prediction

Let $f(x)=x^2$ on $[2,2.1]$, and use $L=4$ as the proposed slope. On this interval, $f'(t)=2t$, so

|f'(t)-4|=|2t-4|\leq0.2 \qquad (2<t<2.1).

The error bound applies with $\varepsilon=0.2$, giving

\left|f(2.1)-f(2)-4(2.1-2)\right| \leq 0.2(0.1) = 0.02.

Direct calculation checks the estimate:

\left|2.1^2-2^2-4(0.1)\right| = |4.41-4-0.4| = 0.01 \leq0.02.

The bound is valid even though the derivative is not exactly $4$ throughout the interval. It uses only a uniform bound on how far the derivative can be from $4$.

Increasing Derivatives and Convexity

Another consequence of the Mean Value Theorem concerns the relative slopes of successive secants. To state it precisely, recall the standard definition: a function $f$ on an interval $I$ is convex if, for any $x,y\in I$ and any $0\leq\lambda\leq1$,

f\bigl(\lambda x+(1-\lambda)y\bigr) \leq \lambda f(x)+(1-\lambda)f(y).

It is strictly convex if the inequality is strict whenever $x\ne y$ and $0<\lambda<1$. Geometrically, convexity means that the function’s graph lies on or below each chord joining two points of the graph; strict convexity places the graph strictly below the chord between distinct points.

Theorem (Increasing Derivative Gives Convexity): Let $I$ be an open interval, and let $f:I\to\mathbb{R}$ be differentiable. If $f'$ is nondecreasing on $I$, then $f$ is convex on $I$. If $f'$ is strictly increasing on $I$, then $f$ is strictly convex on $I$.

Proof. Fix $x<y$ in $I$ and choose $z$ with $x<z<y$. The Mean Value Theorem applied on $[x,z]$ gives a point $u\in(x,z)$ such that

\frac{f(z)-f(x)}{z-x}=f'(u).

Applied on $[z,y]$, the theorem gives a point $v\in(z,y)$ such that

\frac{f(y)-f(z)}{y-z}=f'(v).

Because $u<z<v$, the assumption that $f'$ is nondecreasing implies $f'(u)\leq f'(v)$. Therefore

\frac{f(z)-f(x)}{z-x} \leq \frac{f(y)-f(z)}{y-z}.

Both denominators are positive, so multiplying through by $(z-x)(y-z)$ and collecting the terms involving $f(z)$ yields

(y-x)f(z)\leq (y-z)f(x)+(z-x)f(y).

Set $\lambda=(y-z)/(y-x)$. Then $0<\lambda<1$ and $z=\lambda x+(1-\lambda)y$. Dividing the last inequality by $y-x>0$ gives

f(z)\leq \lambda f(x)+(1-\lambda)f(y).

This proves the convexity inequality for every point strictly between $x$ and $y$. At either endpoint the inequality is equality, so the definition holds for all $0\leq\lambda\leq1$. If $f'$ is strictly increasing, then $u<v$ implies $f'(u)<f'(v)$. The secant-slope inequality and the resulting chord inequality are then strict for every $x<z<y$, proving strict convexity. $\square$

Worked Example: Strict Convexity of a Fourth Power

Let $f(x)=x^4$ on $\mathbb{R}$. Its derivative is $f'(x)=4x^3$. If $x<y$, then

$$ f'(y)-f'(x) = 4(y^3-x^3) = 4(y-x)(y^2+xy+x^2). $$

Here $y-x>0$. Also $y^2+xy+x^2>0$ for $x<y$: it equals $(x+y/2)^2+3y^2/4$, and it can be zero only if $x=y=0$, which is impossible when $x<y$. Thus $f'(y)>f'(x)$, so $f'$ is strictly increasing. The theorem proves that $x^4$ is strictly convex. For this example, $f'$ is in fact strictly increasing, so $x^4$ is strictly convex. The theorem as stated only concludes convexity when the derivative is nondecreasing; its strict-increase hypothesis gives the stronger strict-convexity conclusion.

What the Theorem Does—and Does Not—Say

The Mean Value Theorem is often used as a bridge: it transfers a statement about derivatives at interior points into a statement about changes in function values. The error estimate above uses that bridge to control a linear prediction. The convexity result uses it twice, on adjacent intervals, to compare their secant slopes.

A common mistake is to guess that the point $c$ must be the midpoint, or that it is unique. Neither claim follows from the theorem. The cubic example showed that an equation for $c$ may have several solutions, some of which are excluded because they lie at an endpoint. Another pitfall is to omit continuity on the closed interval or differentiability on the open interval: both hypotheses are needed to apply the theorem to the entire interval. In particular, differentiability at the endpoints is not required.

The theorem guarantees existence, not a formula for the point. In applications, one often does not need to identify $c$: knowing that it exists is enough to establish an inequality or compare slopes. When an explicit point is needed, as in the worked examples, solve the equation $f'(c)=(f(b)-f(a))/(b-a)$ and then check that the solution lies in $(a,b)$.

Check Your Understanding

Use the Mean Value Theorem and its consequences to answer the following.

For $f(x)=x^2$ on $[0,4]$, calculate the average rate of change and find an interior point where the derivative equals it.
Why does a solution of $f'(c)=(f(b)-f(a))/(b-a)$ at $c=b$ not satisfy the theorem’s conclusion?
If $|f'(t)-L|\leq\varepsilon$ on $(a,b)$, what bound does the error estimate give for $f(b)-f(a)-L(b-a)$?
Explain why a strictly increasing derivative gives a strict inequality between the slopes of the secants on $[x,z]$ and $[z,y]$, when $x<z<y$.
For $f(x)=x^4$, verify that $f'$ is strictly increasing and state what the convexity theorem then implies.

What You'll Learn