Answer
(a) $f^{-1}(x)=3-x^2$, $x\geq 0$
(b) See graphs
Work Step by Step
(a)
$f(x)=\sqrt{3-x}$ where $3-x\geq 0$
$y=\sqrt{3-x}$ (Take the square)
$y^2=3-x$ where $y\geq 0$
$x=3-y^2$ (Replace $x$ by $f^{-1}(x)$ and $y$ by $x$)
$f^{-1}(x)=3-x^2$ where $x\geq 0$
We get $f^{-1}(x)=3-x^2$, $x\geq 0$.
(b) See graph
