Mathematics - Application of Derivative Question with Solution | TestHub

Given that, $f(x)=\{\begin{array}{l}(2+x{)}^3;-3<x\le -1\\ x^{2/3};-1<x<2\end{array}$ $$\Rightarrow f^{\prime }(x)=\{\begin{array}{l}3(2+x{)}^2;-3<x\le 1\\ \frac{2}{3}{x}^{-\frac{1}{3}};-1<x<2\end{array}$$
Clearly, $f^{\prime }(x)$ changes its sign at $x=-1$ from $+$ ve to -ve and so $f(x)$ has local maxima at $x=-1$. Also, $f^{\prime }(0)$ does not exist but $f^{\prime }\left(0^{-}\right)<0$ and $f^{\prime }\left(0^{+}\right)<0$. It can only be inferred that $f(x)$ has a possibility of a minima at $x=0$. Hence, the given function has one local maxima at $x=-1$ and one local minima at $x=0$.