How to find a formula for an inverse function

Remember that  $y=f(x)$ means the exact same thing as $x=f^{-1}(y)$. To find a formula for $f^{-1}$,

Warning: in general, $f^{-1}(x)$ is not the same as $f(x)^{-1}$! Inverses and reciprocals are different things.

Example 2: Find a formula for an inverse to $f(x) = 2x^2 + 5$.

Solution: In this case, $f$ does not pass the horizontal line test, so $f$ does not have an inverse defined on the whole domain. Let's restrict the domain to $[0,\infty)$, for example, so that now $f$ does pass the HLT.  DO:  Why does $f$ now pass the HLT?  The range of $f$ is $[5,\infty)$.  Why?  To find a formula for $f^{-1}$, we write: \begin{eqnarray} y&=&f(x) \Leftrightarrow x=f^{-1}(y) )\cr 2x^2 + 5 &=&y\cr x^2 &=& (y-5)/2 \cr x&=&\pm\sqrt{\frac{y-5}{2}}=f^{-1}(y). \end{eqnarray} But here, $x$ is not a function.  So we must determine which square root we want.  Since the range of $f^{-1}$ is the domain of $f$, our answer must be in $[0,\infty)$, so we want the positive root: $f^{-1}(y) = x=+ \sqrt{(y-5)/2}$. This makes sense whenever $y \ge 5$. The domain of $f^{-1}$ is $[5,\infty)$, which is the range of $f$.  At this point, if $x$ and $y$ have no applicable meaning, you could switch $x$ and $y$ if needed to get $y=f^{-1}(x)=\sqrt{\frac{x-5}{2}}$.