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Rolle’s Theorem

Theorem

Let a<ba < b be real numbers, and consider a function f ⁣:[a,b]→Rf\colon [a, b] \to \mathbb{R} that is continuous on the interval [a,b][a, b] and differentiable on (a,b).(a, b)\text{.} If we have f(a)=f(b),f(a) = f(b)\text{,} then there exists a c∈(a,b)c\in (a, b) for which f′(c)=0.f'(c) = 0\text{.}

Interpretation

Rolle’s theorem guarantees a stationary point between two points where a function takes the same value, f(a)=f(b)f(a)=f(b): a graph without breaks or corners that leaves a height and comes back to it must be flat somewhere in between.

A parabola with equal values at the endpoints a and b and a horizontal tangent at a point c between them, where the derivative is 0.

The parabola takes the same value at aa and bb, and the horizontal tangent sits at its lowest point cc. This is how the proof goes. By the extreme value theorem, ff attains a maximum and a minimum on [a,b][a,b]. If both occur at the endpoints, ff is constant and every cc works; otherwise one of them lies at an interior point cc, and there f′(c)=0f'(c)=0 by the first-order condition.

Pick aa; for the presets, the figure finds the next bb in the window with f(b)=f(a)f(b)=f(a) and the c∈(a,b)c\in(a,b) with f′(c)=0f'(c)=0. Custom formulas show numerical candidates without completeness or smoothness guarantees. The tent function is not differentiable at 22, and no horizontal tangent exists.

Each hypothesis closes a gap. Without differentiability, ∣x∣|x| on [−1,1][-1,1] has equal endpoint values but a corner at 00 instead of a horizontal tangent. Without continuity at an endpoint, f(x)=xf(x)=x on [0,1)[0,1) with f(1)=0f(1)=0 has f(0)=f(1)f(0)=f(1), while f′(x)=1f'(x)=1 throughout (0,1)(0,1). The point cc need not be unique: on [0,2π][0,2\pi] the sine has two such points, π/2\pi/2 and 3π/23\pi/2.

The mean value theorem follows by applying Rolle’s theorem to ff minus its secant line. Rolle’s theorem also limits the number of zeros: between two zeros of a differentiable ff lies a zero of f′f', so if f′f' has no zero on an interval, ff has at most one there. For example, f(x)=x3+x−1f(x)=x^3+x-1 has f′(x)=3x2+1>0f'(x)=3x^2+1>0, so it has at most one zero, and the intermediate value theorem gives exactly one, in (0,1)(0,1).

Key Results

  • Continuity and differentiability are essential hypotheses.
  • The conclusion guarantees existence but not uniqueness.
  • Rolle’s theorem is a special case of the mean value theorem.

Examples

  • The function f(x)=x2−1f(x)=x^2-1 on [−1,1][-1,1] satisfies Rolle’s theorem.
  • The sine function satisfies the hypotheses on suitable intervals.

Prerequisites

Sources

Symbols: element of, pi, real numbers

Last updated: 27 October 2026 11:52