Exercise · ZW-MEAN-VALUE-THEOREM-008 · ★★★ Advanced
Find the value c guaranteed by the mean value theorem for \(f(x)=x^{3}-3x\) on \([0,2]\).
Choose one option.
Show answer
- A. \(\frac{2\sqrt{3}}{3}\)
Exercise · ZW-MEAN-VALUE-THEOREM-008 · ★★★ Advanced
Choose one option.
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 \((0,2)\).
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 \((b,f(b))\).