Calculus: Early Transcendentals (2nd Edition)

$f(x)=\sqrt{3-x}$ $f^{-1}(x)=3-x^{2}$
$f(x)=\sqrt{3-x}$, for $x\le3$ Substitute $f(x)$ by $y$: $y=\sqrt{3-x}$ Solve the equation for $x$. Start by squaring both sides: $y^{2}=(\sqrt{3-x})^{2}$ $y^{2}=3-x$ Take $x$ to the left side and $y^{2}$ to the right side: $x=3-y^{2}$ Interchange $x$ and $y$: $y=3-x^{2}$ Substitute $y$ by $f^{-1}(x)$: $f^{-1}(x)=3-x^{2}$ The graph of both the original function and its inverse is shown in the answer section.