Chapter 2 - Section 2.8 - One-to-One Functions and Their Inverses - 2.8 Exercises - Page 227: 61

$f^{-1}(x)=\sqrt{4-x}$

$f(x)=4-x^{2}$ $,$ $x\ge0$ Rewrite this expression as $y=4-x^{2}$ and solve for $x$: $y=4-x^{2}$ Take $-x^{2}$ to the left side and $y$ to the right side: $x^{2}=4-y$ Take the square root of both sides: $\sqrt{x^{2}}=\sqrt{4-y}$ $x=\sqrt{4-y}$ Interchange $x$ and $y$: $y=\sqrt{4-x}$ The inverse of the original function is $f^{-1}(x)=\sqrt{4-x}$