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.
Show answer
- A. \(c=\tfrac94\)
Exercise · ZW-MEAN-VALUE-THEOREM-013 · ★☆☆ Basic
Choose one option.
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.
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 \((b,f(b))\).