91Ó°ÊÓ

Graph of the function

Look at the graph of the function

In the graph we can see that When $f\left(0\right)=-1$, the function crosses the y-axis at -1. So,$f\left(0\right)=-1$

Look at the graph of the function

In the graph we can see that When $x=\mathrm{Ï€}$then y- value of the function is 1 So, $f\left(\mathrm{Ï€}\right)=1$

Look at the graph of the function

In the graph we can see that When $x=-\mathrm{Ï€}$then y-value of the function is 1 So, $f(-\mathrm{Ï€})=-1$

Look at the graph

We observe that graph of the function has $x$value 0 at points $-\frac{3\mathrm{Ï€}}{2},-\frac{\mathrm{Ï€}}{2},\frac{\mathrm{Ï€}}{2},\frac{3\mathrm{Ï€}}{2}$

So the values of $x=-\frac{3\mathrm{Ï€}}{2},-\frac{\mathrm{Ï€}}{2},\frac{\mathrm{Ï€}}{2},\frac{3\mathrm{Ï€}}{2}$

The x-intercepts occur when$y=0$

So the x-intercepts occur when$f\left(x\right)=0$

The x-intercepts are$\left(-\frac{3\mathrm{Ï€}}{2},0\right)\left(-\frac{\mathrm{Ï€}}{2},0\right)\left(\frac{\mathrm{Ï€}}{2},0\right)\left(\frac{3\mathrm{Ï€}}{2},0\right)$

The y-intercepts occur when$x=0$

So y-intercept is$\left(0,-1\right)$

This function has a definite value when $x$is from $-2\mathrm{Ï€}$to $2\mathrm{Ï€}$

Therefore, the domain in interval notation is

This function has a definite value when $y$is from $-1$to$1$

Therefore, the domain in interval notation is $\left[-1,1\right]$