Rolle's theorem

In calculus and real analysis, Rolle's theorem (or lemma) states that a real-valued differentiable function which attains equal values at two distinct points must have a stationary point somewhere between them, that is, a point where its derivative is zero. The theorem is named after Michel Rolle. The theorem is a special case of, and is used to prove, the mean value theorem.

Statement If a real function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one c in the open interval (a, b) such that f ′ ( c ) = 0. {\displaystyle f'(c)=0.}

History In the 12th century Bhāskara II states an early form of Rolle's theorem in his work, though it was stated without a modern formal proof. Although the theorem is named after Michel Rolle, Rolle's 1691 proof covered only the case of polynomial functions. His proof did not use the methods of differential calculus, which at that point in his life he considered to be fallacious. The theorem was first proved by Cauchy in 1823 as a corollary of a proof of the mean value theorem. The name "Rolle's theorem" was first used by Moritz Wilhelm Drobisch of Germany in 1834 and by Giusto Bellavitis of Italy in 1846.

Examples and counterexamples

Differentiability is not needed at the endpoints: Half circle For a positive real number r, consider the function f : [ − r , r ] → R {\displaystyle f:[-r,r]\to \mathbb {R} } such that

f ( x ) = r 2 − x 2 {\displaystyle f(x)={\sqrt {r^{2}-x^{2}}}} for all x ∈ [ − r , r ] . {\displaystyle x\in [-r,r].}

Its graph is the upper semicircle centered at the origin. This function is continuous on the closed interval [−r, r] and differentiable in the open interval (−r, r), but not differentiable at the endpoints −r and r, as the graph of f has vertical tangents at those points. Since f(−r) = f(r), Rolle's theorem applies, and indeed, there is a point where the derivative of f is zero. The theorem applies even when the function cannot be differentiated at the endpoints because it only requires the function to be differentiable in the open interval.

Differentiability is needed within the open interval: Absolute value

If differentiability fails at an interior point of the interval, the conclusion of Rolle's theorem may not hold. Consider the absolute value function

f ( x ) = | x | , x ∈ [ − 1 , 1 ] . {\displaystyle f(x)=|x|,\quad x\in [-1,1].}

Then f(−1) = f(1), but there is no c between −1 and 1 for which the f′(c) is zero. This is because that function, although continuous, is not differentiable at x = 0. The derivative of f changes its sign at x = 0, but without attaining the value 0. The theorem cannot be applied to this function because it does not satisfy the condition that the function must be differentiable for every x in the open interval.

Functions with zero derivative Rolle's theorem implies that a differentiable function whose derivative is ⁠ 0 {\displaystyle 0} ⁠ in an interval is constant in this interval. Indeed, if a and b are two points in an interval where a function f is differentiable, then the function

g ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) b − a ( x − a ) {\displaystyle g(x)=f(x)-f(a)-{\frac {f(b)-f(a)}{b-a}}(x-a)}

satisfies the hypotheses of Rolle's theorem on the interval ⁠ [ a , b ] {\displaystyle [a,b]} ⁠. If the derivative of ⁠ f {\displaystyle f} ⁠ is zero everywhere, the derivative of ⁠ g {\displaystyle g} ⁠ is

g ′ ( x ) = − f ( b ) − f ( a ) b − a , {\displaystyle g'(x)=-{\frac {f(b)-f(a)}{b-a}},}

and Rolle's theorem implies that there is ⁠ c ∈ ( a , b ) {\displaystyle c\in (a,b)} ⁠ such that

0 = g ′ ( c ) = − f ( b ) − f ( a ) b − a . {\displaystyle 0=g'(c)=-{\frac {f(b)-f(a)}{b-a}}.}

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