The total number of local maxima and local minima of the function...

Given that, f(x)=\left\{$\begin{array}{l}(2+x)^3 ;-3 < x \leq-1 \\ x^{2 / 3} ; \quad-1 < x < 2\end{array}$\right. \Rightarrow \quad f^{\prime}(x)=\left\{\begin{array}{l} 3(2+x)^2 ;-3 < x \leq 1 \\ \frac{2}{3} x^{-\frac{1}{3}} ; \quad-1 < x < 2 \end{array}\right.
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 $x=0$.