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Explain why the absence of a zero does not violate Rolle’s theorem

by RoRi

Consider the function

    \[ f(x) = 1 - x^{\frac{2}{3}}. \]

Show that f(1) = 0 and f(-1) = 0, but that the derivative f'(x) \neq 0 for all x \in [-1,1]. Explain why this does not violate Rolle’s theorem.

Proof. First, we show that f(1) = 0 and f(-1) = 0 by a direct computation:

    \begin{align*} f(1) = 1 - (1)^{\frac{2}{3}} = 1 - 1 &= 0, \\ f(-1) = 1 - (-1)^{\frac{2}{3}} = 1 - (1^2)^\frac{1}{3} = 1 - 1 &= 0. \end{align*}

Then, we compute the derivative,

    \[ f(x) = 1 - x^{\frac{2}{3}} \quad \implies \quad f'(x) = -\frac{2}{3} x^{-\frac{1}{3}}. \]

To show f'(x) \neq 0 for any x \in [-1,1] we consider three cases:

  • If x < 0 then x^{-\frac{1}{3}} < 0 implies f'(x) > 0 (since -\frac{2}{3} times a negative is positive).
  • If x > 0 then x^{-\frac{1}{3}} > 0 implies f'(x) < 0 (since -\frac{2}{3} times a positive is then negative).
  • If x = 0, then f'(x) is undefined (since x^{-\frac{1}{3}} = \frac{1}{x^{1/3}}).

Thus, f'(x) \neq 0 for any x \in [-1,1]. \qquad \blacksquare

This is not a violation of Rolle’s theorem since the theorem requires that f(x) be differentiable for all x on the open interval (-1,1). Since f'(x) is not defined at x = 0, we have f(x) is not differentiable on the whole interval. Hence, Rolle’s theorem does not apply.

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