Question

Find the domain of definition of the following function.
$y=\frac{1}{x^3+x-2}.$

Answer 1

$y=\frac{1}{x^3+x-2}$

$x^3+x-2\ne 0$

$x^3+x-2=0$

$x^3+x=2$

$x(x^2+1)=2$

$x=1$

$x^2+1=2$

$x^2=2-1$

$x^2=1$

$x=\pm 1$

When $x=-1$

$y=\frac{1}{(-1{)}^3+(-1)-2}$

$=\frac{1}{-1-1-2}$

$=\frac{1}{-4}$

It is possible

When $x=1$

$y=\frac{1}{(1{)}^3+1-2}$

$\frac{1}{2-2}$

$=\frac{1}{0}$

Which is not defined

So, the domain of $y=(-\infty ,\infty )-\left\{1\right\}$