Let f(x) = ≤ft(4-x^2)^2 / 3, then f has a

$f(x)=\left(4-x^2\right)^{2 / 3}$

$f'(x)=\frac{-4 x}{3\left(4-x^2\right)^{1 / 3}}=\frac{4 x}{3\left(x^2-4\right)^{1 / 3}}$

at x = –2   local minima

x = 0         local maxima

x = 2         local minima