Question

Find the domain of definition of the following function.
$y=\frac{\sqrt{x+12-x^2}}{x^2-4}$

Answer 1

$y=\frac{\sqrt{x+12-x^2}}{x^2-4}$

$x^2-4\ne 0$

$x\ne -2,2$

$-x^2+x+12\ge 0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-1\pm \sqrt{1-4\times -1\times 12}}{2\times -1}$

$\Rightarrow x=\frac{-1\pm \sqrt{49}}{-2}$

$\Rightarrow x=\frac{-1\pm 7}{-2}$

$\Rightarrow x=\frac{-8}{-2},\frac{-1+7}{-2}$

$\Rightarrow x=4,-3$

So, $x\le 4$

$x\ge -3$

When, $x>4$

$x+12-x^2<0$

So, the domain of $f\left(x\right)$ is $[-3,4]-\{2,-2\}$