Question

Find range and domain of $f\left(x\right)=\frac{1}{\sqrt{16-x^2}}$

Answer 1

The function is only defined only if $16-x^2>016>x^2|x|<4x\in (-4,4)$

let $y=f\left(x\right)⟹y=\frac{1}{\sqrt{16-x^2}}\sqrt{16-x^2}=\frac{1}{y}16-x^2=\frac{1}{y^2}{x}^2=16-\frac{1}{y^2}x=\sqrt{16-\frac{1}{y^2}}$

This function is defined only if $16-\frac{1}{y^2}\ge 016y^2-1\ge 0y^2\ge \frac{1}{16}|y|\ge \frac{1}{4}y\in [-4,4]$

So the domain is $(-4,4)$ nd range is $[-4,4]$