Exercise · ZW-ROLLES-THEOREM-016 · ★★☆ Intermediate

For \(f(x)=\sqrt{1-x^2}\) on \([-1,1]\), does Rolle's theorem apply and which point does it give?

Choose one option.

Show answer
  • B. Yes, with \(c=0\), even though \(f\) is not differentiable at \(\pm1\).

Explanation

It is continuous on \([-1,1]\), differentiable on \((-1,1)\) and \(f(\pm1)=0\). Differentiability is only required inside; \(f'(x)=-x/\sqrt{1-x^2}\) vanishes at \(0\).

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Review the concept: Rolle's theorem

If \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\) and \(f(a)=f(b)\), there is at least one \(c\in(a,b)\) with \(f^{\prime}(c)=0\).

\[f(a)=f(b)\ \Rightarrow\ \exists\,c\in(a,b):\ f'(c)=0\]

More practice problems: rolle's theorem

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