Question: **$\{.\}$ represents fractional part function**

$$f’(x)=3x^2-3=0$$
$$\implies x=\pm 1$$

Function is maximum at $x=-1$ and minimum at $x=1$

$$f(-1)=2$$ and $$f(1) =-1$$

One root lies between $-1$ and $1$ and the others lie before and after the numbers respectively.

$$\{x\_1\}+\{x\_2\}+\{x\_3\}$$
$$=(x\_1+x\_2+x\_3)-([x\_1]+[x\_2]+[x\_3])$$
$$3-([x\_1]+0+[x\_3])$$

I don’t know how far $x\_1$ and $x\_3$ are from $-1$ and $1$, so the value of their greatest integer can’t be determined. How do I proceed?

Answer: 
by trial and error and intermediate value theorem

**Observe** $f(0),f(1)$,,

$f(-1),f(-2)$,,

$f(1),f(2)$ are of opposite signs.

thus we can find the intervals of roots.

Let $x\_1=a$, $x\_2=-2+b$, $x\_3=1+c$

but

$x\_1+x\_2+x\_3=0$

or :$a+b+c=1$

Note: $a,b,c$denote the fractional part of $x\_1,x\_2,x\_3$
Easy to see that: $x\_1\in(-2,-1),$ $x\_2\in(0,1)$ and $x\_3\in(1,2).$

Thus, since $\{x\}=x-[x],$ we obtain:
$$\{x\_1\}+\{x\_2\}+\{x\_3\}=2+x\_1+x\_2+x\_3-1=1.$$