Exercise · ZW-ROLLES-THEOREM-017 · ★★☆ Intermediate

Let \(f(x)=x\) if \(0\le x<1\) and \(f(1)=0\). Why does Rolle's conclusion fail on \([0,1]\)?

Choose one option.

Show answer
  • D. Because \(f\) is not continuous on \([0,1]\) (it fails at \(x=1\)).

Explanation

Here \(f(0)=f(1)=0\) and \(f\) is differentiable on \((0,1)\) with \(f'=1\), but there is a jump at \(x=1\): without continuity on the closed interval the theorem does not apply.

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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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