Exercise · ZW-ROLLES-THEOREM-007 · ★★☆ Intermediate

For \(f(x)=x^{2}\) on \([0,2]\), which \(c\in(0,2)\) satisfies \(f'(c)=\frac{f(2)-f(0)}{2-0}\)?

Choose one option.

Show answer
  • D. \(1\)

Explanation

Here \(f(0)\neq f(2)\), so Rolle does not apply; this is the mean value theorem. The secant slope is \(\frac{4-0}{2}=2\), and \(2c=2\) gives \(c=1\).

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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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