Exercise · ZW-ROLLES-THEOREM-015 · ★★☆ Intermediate

For \(f(x)=x^3-3x\) on \([-\sqrt3,\sqrt3]\), which values of \(c\) satisfy \(f'(c)=0\)?

Choose one option.

Show answer
  • A. \(c=-1\) and \(c=1\)

Explanation

The endpoints give \(f(\pm\sqrt3)=0\). Since \(f'(x)=3x^2-3=0\) at \(\pm1\), both points work: the theorem guarantees at least one, not exactly one.

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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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