Problem 961

Solution. We will start from the graph of the function $y=x^{2}$. In the system $x_{1} O_{1} y$, we construct the graph of $y=x_{1}^{2}$. We translate the axis $O_{1} x_{1}$ by the vector $\bar{r}_{1}(0 ; 1)$, obtaining the new system $x_{2} O_{2} y$ (Fig. 34). In this system, we will have the graph of the function $y=x_{2}^{2}-1$. In the system $x_{2} O_{2} y$, we construct the graph of $y=\left|x_{2}^{2}-1\right|$. Now, we translate the axis $O_{2} x_{2}$ by the vector $\bar{r}_{2}(0 ; 2)$, obtaining the system $x O y$. In this system, we will have the graph of the function $y=\left|x^{2}-1\right|-2$. Finally, in the system $x O y$, we construct the graph of the original function (solid line).

We find the points of extremum. $x_{i}=-1$ and $x_{2}=1$ are points of maximum. The points of minimum lie on the axis $O x$, so $y=0$, i.e., $|x^{2}-1|-2=0$, from which $x_{3}=-\sqrt{3}$ and $x_{4}=\sqrt{3}$ are points of minimum. Another point $x=0$ is a point of minimum.

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Fig. 31

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Fig. 32

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Fig. 33

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Fig. 34
We find the intervals of monotonicity: on the intervals $(-\infty ;-\sqrt{3})$, $(-1 ; 0)$, and $(1 ; \sqrt{3})$ the function is decreasing, and on the intervals $(-\sqrt{3} ;-1)$, $(0 ; 1)$, and $(\sqrt{3} ;+\infty)$ the function is increasing.