Question

Find the domain of the following function.
$f\left(x\right)=\frac{\sqrt{x^2+x-6}}{x^2-4}$

Answer 1

$f\left(x\right)=\frac{\sqrt{x^2+x-6}}{x^2-4}$

$(x^2+x-6)\ge 0$

$(x^2-4)\ne 0$

$x\ne -2,2$

$x^2+x-6\ge 0$

$x^2+3x-2x-6\ge 0$

$x(x+3)-2(x+3)\ge 0$

$x=2,x=-3$

When, $x>2$

$x^2+x-6<0$

When, $x<-3$

$x^2+x-6<0$

$x=2$ is not possible.

So, the domain of $f\left(x\right)=(-\alpha ,-3]\cup (2,\infty )$