Let f(x) = $\left(4-x^2\right)^{2 / 3}$, then f has a

a local maxima at x = 0 a local maxima at x = 2 a local maxima at x = –2 none of these

a local maxima at x = 0

$f(x)=\left(4-x^2\right)^{2 / 3}$ $f'(x)=\frac{-4 x}{3\left(4-x^2\right)^{1 / 3}}=\frac{4 x}{3\left(x^2-4\right)^{1 / 3}}$ at x = –2   local minima x = 0         local maxima x = 2         local minima