Question

Find the domain of definition of the following function.
$f\left(x\right)=\frac{\sqrt{12+x-x^2}}{x(x-2)}$

Answer 1

$f\left(x\right)=\frac{\sqrt{12+x-x^2}}{x(x-2)}$

$x(x-2)\ne 0$

$x\ne 0and2$

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

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

$-x^2+4x-3x+12\ge 0$

$-x(x-4)-3(x-4)\ge 0$

$(x+3)(x-4)\le 0$

$\Rightarrow x\in [-3,4]$

So,the Domain of $y=[-3,4]-\{0,2\}$