Mean Value Theorem Proof Workshop

From the Theorem to a Proof Strategy

The Mean Value Theorem turns an average rate of change over an interval into an instantaneous rate of change at some interior point. Its proof, developed earlier in this course, uses an endpoint-matching affine function and Rolle’s Theorem. In this workshop, we take that theorem as established and focus on how to organize arguments that use it: which interval to choose, what equation the theorem supplies, and what the resulting witness does—and does not—tell us.

For $a<b$, if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, the Mean Value Theorem gives a point $c\in(a,b)$ satisfying

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

The quotient on the right is the secant slope across the whole interval. The point $c$ is guaranteed to lie strictly between the endpoints, but the theorem does not generally say that this point is unique. Nor does the theorem say that a particular guessed point works: the derivative must actually equal the secant slope there.

A reliable proof follows a short sequence. First identify the interval and the function. Then verify continuity on the closed interval and differentiability on its interior. Calculate the secant slope carefully. Finally, use the conclusion to answer the specific question—existence, uniqueness, comparison, or a contradiction. The theorem supplies a point, not a complete argument for every claim that might follow.

Fix the interval.
Write down the endpoints $a<b$, including whether any proposed witness is actually interior.

Check the hypotheses.
Confirm continuity on $[a,b]$ and differentiability at every point of $(a,b)$.

Compute the target slope.
Evaluate $\bigl(f(b)-f(a)\bigr)/(b-a)$, keeping the order of the endpoints consistent.

Use the witness precisely.
The theorem gives at least one interior $c$ with $f'(c)$ equal to that slope. Additional claims require additional reasoning.

When the Mean Value Point Is Unique

The Mean Value Theorem guarantees at least one point, but there may be more than one. A useful extra hypothesis for uniqueness is strict monotonicity of the derivative. Strictly increasing or strictly decreasing functions take each value at most once; therefore a prescribed secant slope can match the derivative at most once. The theorem supplies existence, and strict monotonicity supplies uniqueness.

Theorem (Uniqueness of the Mean Value Point): Let $a<b$, and let $f:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $f'$ is strictly monotone on $(a,b)$, then there is exactly one $c\in(a,b)$ such that $$ f'(c)=\frac{f(b)-f(a)}{b-a}. $$

Proof. The hypotheses of the Mean Value Theorem hold, so there is at least one $c\in(a,b)$ satisfying the displayed equation. Suppose $c_1,c_2\in(a,b)$ both satisfy it. Then

f'(c_1)=\frac{f(b)-f(a)}{b-a}=f'(c_2).

If $f'$ is strictly increasing, $c_1<c_2$ would imply $f'(c_1)<f'(c_2)$, and $c_2<c_1$ would imply $f'(c_2)<f'(c_1)$. Both possibilities contradict the equality, so $c_1=c_2$. If $f'$ is strictly decreasing, the corresponding strict inequalities reverse, and again distinct points cannot have equal derivative values. Thus at most one such point exists. Together with existence, this proves that exactly one point satisfies the equation. $\square$

Worked Example: A Secant Slope with One Interior Witness

Let $f(x)=x^2$ on $[-1,3]$. This polynomial is continuous on $[-1,3]$ and differentiable on $(-1,3)$. Its secant slope is

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

Since $f'(x)=2x$, a Mean Value point must satisfy $2c=2$, so $c=1$. This point is in $(-1,3)$, and direct substitution verifies the equation $f'(1)=2$. Moreover, $f'$ is strictly increasing: if $x_1<x_2$, then $2x_1<2x_2$. The uniqueness theorem therefore confirms that $c=1$ is the only Mean Value point on this interval.

Strict monotonicity is a sufficient condition for uniqueness, not a necessary one. A derivative may fail to be strictly monotone on the whole interval while the chosen secant slope is still attained at only one interior point. Conversely, if the derivative takes the secant-slope value at two distinct interior points, the Mean Value point is not unique.

Worked Example: Two Distinct Mean Value Points

Let $f(x)=x^3$ on $[-2,2]$. The function is continuous on the closed interval and differentiable on its interior. Its secant slope is

\frac{f(2)-f(-2)}{2-(-2)} =\frac{8-(-8)}{4} =4.

The derivative is $f'(x)=3x^2$. Solving $3c^2=4$ gives $c=2/\sqrt{3}$ or $c=-2/\sqrt{3}$. Both are interior points because $2/\sqrt{3}<2$, which follows by squaring the positive quantities to get $4/3<4$. At either point,

f'\left(\frac{2}{\sqrt{3}}\right) =3\left(\frac{2}{\sqrt{3}}\right)^2 =3\cdot\frac{4}{3} =4, \qquad f'\left(-\frac{2}{\sqrt{3}}\right) =3\left(-\frac{2}{\sqrt{3}}\right)^2 =3\cdot\frac{4}{3} =4.

Thus each point satisfies the Mean Value equation, and the witness is not unique. The derivative $3x^2$ is not strictly monotone on $(-2,2)$, so the uniqueness theorem does not apply. The theorem’s existence conclusion still applies, but it does not rule out multiple witnesses.

Comparing Secant Slopes on Adjacent Intervals

A second useful proof technique is to apply the Mean Value Theorem separately on two adjacent intervals. If the derivative is strictly increasing, the resulting interior points are ordered, so their derivative values—and hence the two secant slopes—are strictly ordered as well. This gives a direct way to compare slopes without expanding a potentially complicated algebraic expression for each one.

Theorem (Adjacent Secant Slopes for a Strictly Increasing Derivative): Let $a<b<d$, and let $f:[a,d]\to\mathbb{R}$ be continuous on $[a,d]$ and differentiable on $(a,d)$. If $f'$ is strictly increasing on $(a,d)$, then $$ \frac{f(b)-f(a)}{b-a} < \frac{f(d)-f(b)}{d-b}. $$

Proof. Apply the Mean Value Theorem to $f$ on $[a,b]$. Its continuity and differentiability hypotheses hold because they hold on the larger interval. There is $u\in(a,b)$ such that

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

Apply the theorem again on $[b,d]$. There is $v\in(b,d)$ such that

\frac{f(d)-f(b)}{d-b}=f'(v).

Because $u<b<v$, we have $u<v$. Strict increase of $f'$ gives $f'(u)<f'(v)$. Substituting the two secant-slope identities yields

\frac{f(b)-f(a)}{b-a} < \frac{f(d)-f(b)}{d-b},

as required. $\square$

Worked Example: Comparing Two Secant Slopes

Take $f(x)=x^2$ and the ordered points $-1<1<3$. The derivative $f'(x)=2x$ is strictly increasing. The secant slope from $-1$ to $1$ is

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

The secant slope from $1$ to $3$ is

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

The theorem predicts that the first slope is less than the second, and the calculations verify $0<4$. The proof of the theorem explains why: the Mean Value point in $(-1,1)$ lies to the left of the one in $(1,3)$, and a strictly increasing derivative has a smaller value at the left-hand point.

Proof Audit: Do Not Skip the Hypotheses

The assumptions on the Mean Value Theorem are not formalities. Continuity is required on the closed interval, while differentiability is required at every interior point. An endpoint need not have a derivative for the theorem to apply; an interior point where differentiability fails can prevent the application. A plausible-looking secant-slope calculation cannot replace these checks.

Worked Example: A Secant Slope the Theorem Cannot Match

Consider $f(x)=|x|$ on $[-1,1]$. It is continuous on this interval, and its secant slope is

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

For $x<0$, $f(x)=-x$, so $f'(x)=-1$. For $x>0$, $f(x)=x$, so $f'(x)=1$. The derivative does not exist at the interior point $0$: the left difference quotient there is $-1$, while the right difference quotient is $1$. In fact, there is no interior point where the derivative equals the secant slope $0$. This does not contradict the Mean Value Theorem, because differentiability fails at $0$, and the theorem’s hypotheses are not satisfied.

When checking a proof, it is also worth distinguishing the point the theorem guarantees from a point that merely looks promising. For example, in the quadratic example on $[-1,3]$, the candidate $c=1$ was not accepted just because it was plausible: it was verified by the equation $f'(1)=2$. In a more complicated problem, solving the derivative equation may produce candidates outside the open interval, or none at all; either outcome must be reconciled with the theorem’s hypotheses and the calculation.

The endpoint-matching affine construction from the proof of the Mean Value Theorem remains useful for understanding why the theorem is true. In applications, however, it is usually more efficient to cite the theorem by name, verify its assumptions, and state exactly how its conclusion is used. The uniqueness theorem and the adjacent-slope comparison show two distinct ways to add information: strict monotonicity can control how many witnesses exist, and ordering the witnesses can compare slopes on separate intervals.

Check Your Understanding

Use the Mean Value Theorem and the results proved above to answer the following questions.

Which hypotheses must be checked before applying the Mean Value Theorem on $[a,b]$, and on which parts of the interval are they required?
Why does strict monotonicity of $f'$ imply uniqueness of a point satisfying the Mean Value equation, but not provide existence by itself?
For $f(x)=x^3$ on $[-2,2]$, what is the secant slope, and which two interior points satisfy the Mean Value equation?
In the adjacent-slope theorem, why is the Mean Value point for $[a,b]$ to the left of the point for $[b,d]$?
Why does $f(x)=|x|$ on $[-1,1]$ not contradict the Mean Value Theorem, despite having no point where its derivative equals its secant slope?