Exercise · ZW-MEAN-VALUE-THEOREM-013 · ★☆☆ Basic

For \(f(x)=\sqrt{x}\) on \([1,4]\), what is the point \(c\) of the mean value theorem?

Choose one option.

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  • A. \(c=\tfrac94\)

Explanation

The chord slope is \(\frac{2-1}{4-1}=\frac13\) and \(f'(c)=\frac{1}{2\sqrt c}=\frac13\) gives \(\sqrt c=\frac32\), so \(c=\frac94\). The midpoint \(\frac52\) does not work in general.

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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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