Exercise · ZW-MEAN-VALUE-THEOREM-002 · ★★☆ Intermediate

Find the value c guaranteed by the mean value theorem for \(f(x)=x^{3}\) on \([0,2]\).

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  • C. \(\frac{2 \sqrt{3}}{3}\)

Explanation

The slope of the secant is \(\dfrac{f(2)-f(0)}{2-0}=4\). Solve \(f'(c)=4\) with \(f'(x)=3 x^{2}\) and the solution inside \((0,2)\) is c = \frac{2 \sqrt{3}}{3}.

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Review the concept: Mean value theorem

If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), there is a \(c\in(a,b)\) such that the slope of the tangent at \(c\) equals the slope of the secant line joining \((a,f(a))\) and \((b,f(b))\).

\[f'(c)=\frac{f(b)-f(a)}{b-a}\]

More practice problems: mean value theorem

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