4 -4 -3 2 4 -2

See from the given graph of the function $f\left(x\right)$, the graph becomes parallel to x axis at the points $x=-2 \text{and} x=1$

Thus, the derivative $f\text{'}\left(x\right)$ becomes zero at $x=-2$ and $x=1$. That is,

$$\begin{gathered} f\text{'}\left(1\right)=0 \\ f\text{'}\left(-2\right)=0 \end{gathered}$$

Thus, $f\text{'}\left(x\right)$ has roots at $x=1 \text{and} x=-2$.

Also see that, $f\left(x\right)$ is an increasing function for $-2<x<1$. Thus,

$f\text{'}\left(x\right)>0, \text{for} -2<x<1$

Also see that, $f\left(x\right)$ is a decreasing  function for $-\infty <x<-2 \text{and} 1<x<\infty$. Thus,

$f\text{'}\left(x\right)<0, \text{for} -\infty <x<-2 \text{and} 1<x<\infty$