Why a Guaranteed Point Exists
Rolle’s Theorem makes a specific promise: if a function begins and ends at the same height, and it is continuous and differentiable in the required places, then its graph has at least one interior point where the tangent is horizontal. The theorem is more than an observation about how a graph might look. Its guarantee follows from two precise facts: a continuous function attains its extreme values on a closed interval, and at an interior extremum a derivative, if it exists, must be zero.
The proof connects these facts. Equal endpoint values alone do not tell us where an interior point can be found. The Extreme Value Theorem supplies an attained maximum or minimum; the endpoint condition ensures that, unless the function is constant, at least one such extreme occurs inside the interval. Differentiability then forces the derivative there to vanish.
The Theorem and Its Conditions
The locations in the hypotheses matter. Continuity is required on the entire closed interval, including its endpoints. Differentiability is required at every point strictly between the endpoints; the theorem does not require derivatives at \(a\) or \(b\). The conclusion is an existence statement: it promises at least one such \(c\), not necessarily exactly one.
The Extreme Value Theorem states that a real-valued function continuous on a closed, bounded interval \([a,b]\) attains both a maximum and a minimum there. In other words, there are points \(x_{\max},x_{\min}\in[a,b]\) such that \(f(x_{\max})\geq f(x)\) and \(f(x_{\min})\leq f(x)\) for every \(x\in[a,b]\). Attaining matters: the maximum and minimum are values of the function at points in the interval, not merely bounds the function approaches.
Why an Interior Extremum Has Derivative Zero
A point \(c\) is a local maximum of \(f\) if there is a positive distance around \(c\) such that \(f(x)\leq f(c)\) for all domain points \(x\) within that distance. A local minimum is defined by reversing the inequality. A local maximum or local minimum is called a local extremum. The following fact applies when the point is interior and the function is differentiable there.
Here is a proof for a local maximum. For every sufficiently small nonzero \(h\), the point \(c+h\) is in the domain and \(f(c+h)\leq f(c)\). If \(h>0\), division by \(h\) preserves the inequality:
If \(h<0\), division reverses the inequality:
Because \(f\) is differentiable at \(c\), the difference quotient has a single two-sided limit as \(h\to 0\), namely \(f'(c)\). The quotients approaching from the right are nonpositive, so their limit is nonpositive. The quotients approaching from the left are nonnegative, so their limit is nonnegative. The same derivative must satisfy both conclusions. Therefore \(f'(c)\leq 0\) and \(f'(c)\geq 0\), which together give \(f'(c)=0\). For a local minimum, the inequalities reverse: right-hand quotients are nonnegative and left-hand quotients are nonpositive. Their common limit is again zero. This proves the lemma for either type of local extremum.
Proof of Rolle’s Theorem
Set \(k=f(a)=f(b)\). There are two cases. Keeping them separate is useful: the constant case already gives the conclusion, while the nonconstant case is where the Extreme Value Theorem identifies an interior extremum.
If \(f(x)=k\) for every \(x\in[a,b]\), then \(f'(x)=0\) at every \(x\in(a,b)\). Since \(a<b\), the open interval contains points, so there is a point \(c\in(a,b)\) with \(f'(c)=0\).
If \(f\) is not constant, there is some \(x_0\in[a,b]\) for which \(f(x_0)\neq k\). Since the endpoint values both equal \(k\), this \(x_0\) must lie in \((a,b)\). If \(f(x_0)>k\), the attained global maximum is greater than \(k\), so it cannot occur at either endpoint. If \(f(x_0)<k\), the attained global minimum is less than \(k\), so it cannot occur at either endpoint. In either situation, an extreme occurs at some \(c\in(a,b)\).
A global maximum or minimum on \([a,b]\) is also a local extremum at the interior point \(c\). The function is differentiable there by hypothesis. The lemma therefore gives \(f'(c)=0\).
The Extreme Value Theorem is applicable in the second step because \(f\) is continuous on \([a,b]\), and the interval is closed and bounded. The resulting extremum is interior because its value is strictly above or below the common endpoint value. This completes the proof in both cases.
Worked Example: A Minimum Between Equal Endpoint Values
Worked Example: A Minimum Between Equal Endpoint Values
Consider \(f(x)=(x-2)^2-1\) on \([1,3]\). First check the endpoint values:
The function is a polynomial, so it is continuous on \([1,3]\) and differentiable on \((1,3)\). The equal-endpoint condition holds. At the interior point \(x=2\),
The point \(2\) is a minimum because \((x-2)^2\geq 0\), so \(f(x)\geq -1=f(2)\) for every \(x\). As a check, the derivative calculation directly gives the required zero, and the endpoint substitutions confirm the theorem’s equal-value condition.
Worked Example: A Maximum Above the Endpoint Values
Worked Example: A Maximum Above the Endpoint Values
Let \(g(x)=-(x-1)(x-3)\) on \([1,3]\). The endpoint values are
The function is a polynomial, so it satisfies the continuity and differentiability conditions. To identify an interior maximum, expand and complete the square:
Since \((x-2)^2\geq 0\), \(g(x)\leq 1\), with equality at \(x=2\). Thus \(x=2\) is an interior maximum. Differentiating and evaluating gives
Substitution into the original expression also checks the value: \(g(2)=-(2-1)(2-3)=1\). The maximum is strictly greater than the two endpoint values, and the derivative at its interior location is zero, as the proof predicts.
Worked Example: The Constant Case
Worked Example: The Constant Case
Take \(h(x)=5\) on \([-2,4]\). Its endpoint values agree:
The function is continuous on the closed interval and differentiable on its interior. Its derivative is
For example, choose the interior point \(c=0\). Then \(-2<0<4\) and \(h'(0)=0\). Direct substitution confirms the constant value \(h(0)=5\); differentiating confirms the zero derivative. This case explains why the proof must allow for a function with no single special interior maximum or minimum: every interior point is both a maximum and a minimum.
A Common Proof Error: Using Only One Side
The derivative at an interior extremum is a two-sided limit. Looking only at difference quotients from the right can establish at most a one-sided restriction; it cannot by itself show that the derivative equals zero. For example, \(q(x)=-x^2\) has a local maximum at \(x=0\), and
For \(h>0\), this quotient is negative. A mistaken argument might stop there and conclude that the derivative is negative. But as \(h\to 0^+\), \(-h\to 0\), not to a negative number. More importantly, for \(h<0\), the quotient \(-h\) is positive, and it also tends to zero. The two-sided limit is therefore
The correction is to use both signs of \(h\), as in the lemma’s proof. The interior condition matters for the same reason: both positive and negative small increments must remain in the domain around the point.
Why the Argument Matters
The proof identifies the distinct work done by each hypothesis. Continuity on the closed interval lets the Extreme Value Theorem produce an attained extreme. Equal endpoint values ensure that a nonconstant function must rise above or fall below that shared value somewhere in the interval, pushing an appropriate extreme into the interior. Differentiability at that interior point then turns the extremum into a zero derivative.
If any link in this chain is omitted, the conclusion has not yet been justified. In particular, the proof does not claim that every function with equal endpoint values has a horizontal tangent merely because its endpoints match. It establishes the guarantee under the stated continuity and differentiability conditions, and shows precisely how those conditions lead to the interior point.
Check Your Understanding
Use the proof’s cases and the interior-extremum lemma to answer these questions.
- Why does the Extreme Value Theorem apply to a function continuous on \([a,b]\)?
- In the nonconstant case, why must a point \(x_0\) with \(f(x_0)\neq f(a)\) lie in the open interval?
- Why does an attained maximum above the common endpoint value have to occur in the interior?
- For a differentiable function with a local maximum at \(c\), what sign does the difference quotient have when \(h>0\), and what sign does it have when \(h<0\)?
- Why is it not sufficient to use only the right-hand difference quotients to conclude that \(f'(c)=0\)?
