Proof of Rolle's Theorem

From Equal Endpoint Values to a Zero Derivative

Rolle’s Theorem, stated earlier in this course, turns an equality at two endpoints into a conclusion about a derivative somewhere between them. Its proof brings together two results already established: the Extreme Value Theorem, which guarantees that a continuous function on a closed interval attains its absolute maximum and minimum, and Fermat’s Theorem, which says that a differentiable function has derivative zero at an interior local extremum.

The main issue is that the equal endpoint values do not, by themselves, tell us where an extremum occurs. If the function is constant, every interior point is already a point with derivative zero. Otherwise, its values differ from the common endpoint value somewhere; the Extreme Value Theorem then forces a strict maximum or minimum to occur in the interior. Fermat’s Theorem completes the argument.

Rolle’s Theorem (stated earlier): Let \(a<b\), and let \(f:[a,b]\to\mathbb{R}\) be continuous on \([a,b]\) and differentiable on \((a,b)\). If \(f(a)=f(b)\), then there exists \(c\in(a,b)\) such that \(f'(c)=0\).

The Interior-Extremum Step

We first make precise the part of the argument that uses equal endpoint values. Let their common value be \(K\). If the function is not constant, it must take a value other than \(K\). A value above \(K\) forces an absolute maximum above both endpoint values; a value below \(K\) forces an absolute minimum below both endpoint values. In either case, the relevant extremum cannot occur at an endpoint.

Lemma (Interior Extremum from Equal Endpoint Values): Let \(a<b\), and suppose \(f:[a,b]\to\mathbb{R}\) is continuous on \([a,b]\), satisfies \(f(a)=f(b)=K\), and is not constant. Then \(f\) has an absolute maximum or an absolute minimum at some point \(c\in(a,b)\).

Proof. Since \(f\) is not constant and both endpoint values equal \(K\), there is a point \(x\in[a,b]\) such that \(f(x)\ne K\). If \(f(x)>K\), the Extreme Value Theorem gives a point \(c\in[a,b]\) at which \(f\) attains its absolute maximum. Its value satisfies

f(c)\geq f(x)>K=f(a)=f(b).

Therefore \(c\) cannot equal \(a\) or \(b\), so \(c\in(a,b)\). If instead \(f(x)<K\), the Extreme Value Theorem gives a point \(c\in[a,b]\) at which \(f\) attains its absolute minimum, and

f(c)\leq f(x)<K=f(a)=f(b).

Again, \(c\) cannot be either endpoint, so \(c\in(a,b)\). These cases cover every possibility for \(f(x)\ne K\), and prove the lemma. \(\square\)

The closed interval in this lemma is important: it allows the Extreme Value Theorem to be applied. The strict inequality is important too: it rules out both endpoints as locations for the selected extremum. The lemma does not require differentiability; that condition enters only when we use Fermat’s Theorem.

Proof of Rolle’s Theorem

We now split the proof into the constant and nonconstant cases. This distinction avoids trying to find a strict interior extremum when there is none: a constant function already has derivative zero at every interior point.

Proof of Rolle’s Theorem: Let \(a<b\), and suppose \(f:[a,b]\to\mathbb{R}\) is continuous on \([a,b]\), differentiable on \((a,b)\), and satisfies \(f(a)=f(b)\). Then there is a point \(c\in(a,b)\) such that \(f'(c)=0\).

Proof. Write \(K=f(a)=f(b)\). If \(f\) is constant on \([a,b]\), then \(f(t)=K\) for every \(t\in[a,b]\). In particular, its derivative at every point of \((a,b)\) is zero. Since \(a<b\), the midpoint \(c=(a+b)/2\) lies in \((a,b)\), and \(f'(c)=0\).

Suppose instead that \(f\) is not constant. The Interior Extremum Lemma gives an absolute maximum or absolute minimum at a point \(c\in(a,b)\). An absolute maximum is a local maximum, and an absolute minimum is a local minimum. Since \(f\) is differentiable at \(c\), Fermat’s Theorem applies and gives \(f'(c)=0\). Thus in either case there is a point \(c\in(a,b)\) with derivative zero, as claimed. \(\square\)

Each hypothesis has a distinct job in this proof. Continuity on the closed interval supplies an attained absolute extremum through the Extreme Value Theorem. Equal endpoint values ensure that, in the nonconstant case, a suitable extremum lies strictly inside the interval. Differentiability at that interior point lets us invoke Fermat’s Theorem. The proof does not require differentiability at the endpoints.

Worked Examples: Applying the Proof

Worked Example: A Quadratic with a Guaranteed Interior Minimum

Let \(f(x)=x^2-2x\) on \([0,2]\). As a polynomial, \(f\) is continuous on \([0,2]\) and differentiable on \((0,2)\). Its endpoint values agree:

f(0)=0^2-2(0)=0, \qquad f(2)=2^2-2(2)=4-4=0.

The function is not constant, so the Interior Extremum Lemma ensures that it has an absolute extremum at an interior point. In this case, the derivative identifies that point:

f'(x)=2x-2, \qquad f'(x)=0 \Longleftrightarrow 2x-2=0 \Longleftrightarrow x=1.

Since \(1\in(0,2)\), this is the point guaranteed by Rolle’s Theorem. Direct substitution gives \(f(1)=1-2=-1\), below the common endpoint value \(0\), so it is the interior absolute minimum.

Worked Example: Several Interior Points with Zero Derivative

Consider \(g(x)=(x-1)^2(x-3)^2\) on \([0,4]\). It is a polynomial, and its endpoint values are

g(0)=(-1)^2(-3)^2=1\cdot9=9, \qquad g(4)=3^2\cdot1^2=9.

The function is not constant. Differentiating the product gives

$$ g'(x)=2(x-1)(x-3)^2+2(x-1)^2(x-3) =4(x-1)(x-2)(x-3). $$

Thus \(g'(x)=0\) at \(x=1\), \(x=2\), and \(x=3\), all of which lie in \((0,4)\). The theorem guarantees at least one such point, not exactly one. In fact, \(g(1)=g(3)=0\), while \(g(2)=1\); the graph has multiple interior points with horizontal tangents.

Worked Example: Equal Endpoint Values Without Differentiability

Define \(h(x)=|2x-1|\) on \([0,1]\). This function is continuous, and its endpoint values are equal:

h(0)=|-1|=1, \qquad h(1)=|1|=1.

However, \(h\) is not differentiable at \(x=1/2\). For \(x<1/2\), \(h(x)=1-2x\), so \(h'(x)=-2\). For \(x>1/2\), \(h(x)=2x-1\), so \(h'(x)=2\). There is no point in \((0,1)\) where the derivative is zero; at the only point where the formula changes, the derivative does not exist. The function has an interior minimum, but the differentiability hypothesis needed for Fermat’s Theorem fails there.

Worked Example: Equal Endpoint Values Without Endpoint Continuity

Define \(q:[0,1]\to\mathbb{R}\) by \(q(x)=x\) for \(0\leq x<1\), and \(q(1)=0\). Then \(q(0)=q(1)=0\), and \(q\) is differentiable at every point of \((0,1)\), with \(q'(x)=1\). But \(q\) is not continuous at \(1\): as \(x\) approaches \(1\) from below, \(q(x)=x\) approaches \(1\), not \(q(1)=0\). The conclusion of Rolle’s Theorem fails, since the derivative is never zero in \((0,1)\). This shows why continuity on the closed interval, including at its endpoints, is part of the hypothesis.

Why the Two Cases Matter

The constant case is not a technical inconvenience. It is the case in which the “find an interior extremum strictly above or below the endpoints” strategy cannot work: all values equal the endpoint value. Instead, constancy gives the conclusion immediately. For a nonconstant function, the lemma provides the strict interior extremum, and Fermat’s Theorem applies there.

A useful way to keep the proof organized is to track which result provides each step:

Check whether the function is constant.
If it is, choose any interior point; its derivative is zero.

Use continuity and equal endpoint values.
For a nonconstant function, the Interior Extremum Lemma gives an absolute extremum at an interior point.

Use differentiability at that point.
Fermat’s Theorem gives derivative zero at the interior extremum.

A common gap in an informal proof is to say that a function “must turn around” without identifying a theorem that guarantees an interior point where this happens. The Extreme Value Theorem and the Interior Extremum Lemma supply that missing step. Another gap is to apply Fermat’s Theorem at an endpoint; Fermat’s Theorem requires an interior extremum, which is why ruling out the endpoints is essential.

Rolle’s Theorem is therefore not just a visual claim about a curve returning to its starting height. The proof depends on a precise chain: continuity on a closed interval gives an attained extremum; nonconstancy and equal endpoint values place an extremum in the interior; differentiability and Fermat’s Theorem force the derivative to vanish there. If any link is missing, the conclusion may fail, as the examples demonstrate.

Check Your Understanding

Use the proof and hypotheses of Rolle’s Theorem to answer the following.

Why does a nonconstant function with equal endpoint values have an absolute extremum in the interior?
Where does the proof use continuity on the closed interval, and where does it use differentiability?
For \(r(x)=x^2-4x\) on \([0,4]\), verify the endpoint values and find a point where the derivative is zero.
Why is the constant-function case handled separately in the proof?
For \(h(x)=|2x-1|\) on \([0,1]\), which hypothesis of Rolle’s Theorem fails, and why does the conclusion fail?

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