Exercise · ZW-MEAN-VALUE-THEOREM-003 · ★★☆ Intermediate

Find the value c guaranteed by the mean value theorem for \(f(x)=\sqrt{x}\) on \([1,4]\).

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  • B. \(\frac{9}{4}\)

Explanation

The slope of the secant is \(\dfrac{f(4)-f(1)}{4-1}=\frac{1}{3}\). Solve \(f'(c)=\frac{1}{3}\) with \(f'(x)=\frac{1}{2 \sqrt{x}}\) and the solution inside \((1,4)\) is c = \frac{9}{4}.

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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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