The total number of local maxima and local minima of the functio $f\left(x\right)=\left\{\begin{array}{cc}(2+x{)}^3& ,-3<x\le -1\\ x^{2/3}& ,-1<x<2\end{array}\right.$

Since $\underset{x\to -1-}{lim} f\left(x\right)=\underset{x\to -1+}{lim} f\left(x\right)=1$

So, $f$ is continuous on (– 3, 2). Also

$f^{\mathrm{\prime }}\left(x\right)=\left\{\begin{array}{cc}3(x+2{)}^2& ,-3<x<-1\\ (2/3)x^{-1/3}& ,-1<x<0,0<x<2\end{array}\right.$

$f\text{'} \left(0\right)$ does not exist

$\begin{array}{l}\mathrm{L}{f}^{\mathrm{\prime }}(-1)=\underset{x\to -1-}{lim} \frac{f\left(x\right)-f\left(1\right)}{x+1}=\underset{x\to -1-}{lim} \frac{(x+2{)}^3-1}{x+1}\\ =\underset{x\to -1-}{lim} \frac{(x+1)\left[(x+2{)}^2+(x+2)+1\right]}{x+1}=3\\ \mathrm{R}{f}^{\mathrm{\prime }}(-1)=\underset{x\to -1+}{lim} \frac{x^{2/3}-1}{x+1}\\ =\underset{x\to 1+}{lim} \frac{\left(x^{1/3}-1\right)\left(x^{1/3}+1\right)}{\left(x^{1/3}+1\right)\left(x^{2/3}+x^{1/3}+1\right)}=-2\end{array}$

So $f$is not differentiable at $x = – 1$

For $x\ne 0,-1,f^{\mathrm{\prime }}\left(x\right)\ne 0\text{. }$ Critical points of $f$ are $-1,0$ At

$x=-1,f$ has a local maximum and $x=0,f$ has a local minimum.