Let's determine the zeros and local extrema of the function

$$
y=f(x)=x^{2}-\{x\}^{2}+x+\left|x^{2}-1\right|
$$

where $\{x\}$ denotes the fractional part of $x$.
Ha $k \leqq x0$. Taking into account what has been said about the jumps, we can conclude that $f(x)$ is strictly increasing for $x \geqq 1$, so there is no zero point here, and since $f(1)=2$.

b) If $k=0$, that is, $0 \leqq x0\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_fe97034ad0a54fd48356g-1.jpg?height=626&width=767&top_left_y=1712&top_left_x=668)

d) If $kf(x)$, if $x$ falls into the $\delta$-neighborhood of the number $k$ and $x \neq k)$.

From the above, it also follows that if $k<-1$, then in the interval $[k, k+1)$, there can only be a zero point if the left-hand limit of $f(x)$ at $(k+1)$ is negative, that is,

$$
\left|(k+1)^{2}-1\right|+(k+1)^{2}+(k+1)-1=k(2 k+5)+1<0
$$

Thus, a zero point can only be in the interval $[-2,-1]$, which we can easily obtain from (3) with the value $k=-2$:

$$
\left(x-\frac{3}{2}\right)^{2}-\frac{29}{4}=0
$$

that is,

$$
x_{2}^{*}=\frac{3-\sqrt{29}}{2}=-1.193
$$

In summary: we have determined that the local maximum points of the function (1) are: $x_{0}=\frac{1}{2}, x_{1}=-\frac{1}{2}, x_{k}=-k(k=$ $2,3, \ldots)$; there is no local minimum point, and it has two zero points, these are: $x_{1}^{*}=-1$ and $x_{2}^{*}=-1.193$.