Exercise · ZW-MEAN-VALUE-THEOREM-008 · ★★★ Advanced

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

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  • A. \(\frac{2\sqrt{3}}{3}\)

Explanation

The secant has slope \(\frac{f(2)-f(0)}{2}=1\). From \(f'(c)=3c^{2}-3=1\) we get \(c^{2}=\frac{4}{3}\), and only \(c=\frac{2\sqrt{3}}{3}\) lies in \((0,2)\).

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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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