Prove Rolle's Theorem

Equal Endpoint Values Force a Stationary Point

The previous tutorial proved that differentiability at a point implies continuity there. That connection lets us combine two ideas: a continuous function on a closed interval attains its maximum and minimum, and a differentiable function has derivative zero at any interior local extremum. When the function has equal values at the two endpoints, one of its extrema must occur inside the interval unless the function is constant.

This argument proves Rolle’s Theorem. It is important to keep its hypotheses together: continuity is required on the entire closed interval, differentiability is required at every interior point, and the endpoint values must agree. The point supplied by the theorem lies strictly between the endpoints.

The proof has two distinct stages. First, the Extreme Value Theorem supplies points where \(f\) attains its largest and smallest values. Second, we show that an appropriate extremum occurs in the interior and use a derivative test there. We establish that derivative test before applying it.

The Derivative at an Interior Extremum

A local maximum at \(c\) means that \(f(x)\leq f(c)\) for all \(x\) sufficiently close to \(c\). A local minimum reverses this inequality. At an interior point, increments \(h\) can be both positive and negative. The difference quotient therefore has opposite sign restrictions on its two sides, and differentiability forces its limit to be zero.

Proof. First suppose \(c\) is a local maximum. There is an \(r>0\) such that \(f(c+h)\leq f(c)\) whenever \(|h|<r\) and \(c+h\) lies in the domain. For \(0<h<r\), dividing \(f(c+h)-f(c)\leq0\) by \(h>0\) gives

For \(-r<h<0\), division by \(h<0\) reverses the inequality, so \(\frac{f(c+h)-f(c)}{h}\geq0\). Because \(f\) is differentiable at \(c\), both one-sided difference quotients approach the same finite limit \(L=f'(c)\).

We verify the signs of that limit. If \(L>0\), the definition of the limit gives a \(\delta>0\) such that \(0<|h|<\delta\) implies \(\left|\frac{f(c+h)-f(c)}h-L\right|<L/2\). For positive \(h\) small enough to satisfy both \(h<r\) and \(h<\delta\), this implies \(\frac{f(c+h)-f(c)}h>L/2>0\), contradicting the nonpositive quotient above. Thus \(L\leq0\). If \(L<0\), choose \(\delta>0\) such that the quotient differs from \(L\) by less than \(|L|/2\) whenever \(0<|h|<\delta\). For a negative \(h\) sufficiently close to zero, the quotient is then less than \(L+|L|/2=L/2<0\), contradicting the nonnegative quotient. Thus \(L\geq0\). Together, \(L=0\).

If \(c\) is a local minimum, apply the maximum argument to \(-f\). The function \(-f\) is differentiable at \(c\), has a local maximum there, and satisfies \((-f)'(c)=-f'(c)\). Hence \(-f'(c)=0\), so \(f'(c)=0\). This proves the lemma. \(\square\)

The interior-point condition matters. At an endpoint, only one direction of increments may be available, so the two opposite sign restrictions need not hold. For example, \(f(x)=x\) has a minimum at the left endpoint \(0\) of \([0,1]\), but \(f'(0)=1\).

Proof of Rolle’s Theorem

Let \(f\) satisfy the hypotheses of Rolle’s Theorem. By the Extreme Value Theorem on a closed interval, there are points \(x_{\min},x_{\max}\in[a,b]\) where \(f\) attains its minimum and maximum. Write \(k=f(a)=f(b)\).

If \(f\) is constant on \([a,b]\), then \(f'(c)=0\) at every \(c\in(a,b)\): for each such \(c\), the difference quotient is zero for every sufficiently small nonzero increment. Since \(a<b\), the interval \((a,b)\) is nonempty, so any point in it gives the conclusion.

Now suppose \(f\) is not constant. There is a point \(x_0\in[a,b]\) such that \(f(x_0)\neq k\). If \(f(x_0)>k\), then the maximum value \(f(x_{\max})\) is at least \(f(x_0)>k\). Neither endpoint can be \(x_{\max}\), because both endpoints have value \(k\). Thus \(x_{\max}\in(a,b)\), and it is an interior local maximum. Fermat’s Theorem gives \(f'(x_{\max})=0\).

If instead \(f(x_0)<k\), then the minimum value \(f(x_{\min})\) is at most \(f(x_0)<k\). Again, neither endpoint can be \(x_{\min}\), so \(x_{\min}\in(a,b)\). It is an interior local minimum, and Fermat’s Theorem gives \(f'(x_{\min})=0\). In either case there is a \(c\in(a,b)\) with \(f'(c)=0\). This proves Rolle’s Theorem. \(\square\)

The constant case must be included: a nonconstant function is not guaranteed, and if the function is constant then every interior point works. In the nonconstant case, the equal endpoint values ensure that any extremum whose value is strictly above or below the endpoint value cannot occur at an endpoint. This is the step that produces an interior point where Fermat’s Theorem applies.

Worked Examples

Why the Hypotheses Matter

Equal endpoint values alone are not sufficient. For \(f(x)=|x|\) on \([-1,1]\), the endpoint values agree and the function is continuous, but it is not differentiable at \(0\). At every interior point other than \(0\), its derivative is either \(1\) or \(-1\), so there is no interior point where its derivative is zero. This does not contradict Rolle’s Theorem: the differentiability hypothesis fails at \(0\).

Continuity on the closed interval is used to invoke the Extreme Value Theorem and obtain attained extrema. Differentiability in the interior is used when applying Fermat’s Theorem. Equal endpoint values are what prevent a strict maximum or minimum value from being attained only at an endpoint. These roles are separate; a proof that omits one of them may not establish the conclusion.

A common proof error is to say that equal endpoint values make an interior maximum or minimum automatic. That is true only after considering whether the function is constant. For a constant function, the extrema occur everywhere, including at the endpoints, but any interior point still has derivative zero. For a nonconstant function, choosing a point whose value differs from the common endpoint value ensures that either the maximum or the minimum is strictly different from that value, and consequently is attained in the interior.