CUET Preparation Today
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 |
