Question

Find the domain and range of $\sqrt{4x-x^2}$.

Answer 1

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

Domains:-$4x-x^2\le 0x(4-x)\le xϵ[0,4]\Rightarrow x^2-4x+a^2=0[a>0]x=\frac{4\pm \sqrt{16-4a^2}}{2}=1\pm \sqrt{1-a^2}\Rightarrow 4-a^2\le 0\Rightarrow aϵ[0,2]=Range$