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 it does not have an inverse defined on the whole domain. Let's restrict the domain to $[0,\infty)$, for example. The range of $f$ is $[5,\infty)$.
To find a formula for $f^{-1}(x)$, we write:
\begin{eqnarray} y&=&f^{-1}(x) \cr x&=&f(y) \cr x&=&2y^2 + 5 \cr y^2
&=& (x-5)/2. \end{eqnarray}
To find $y$ we need to take a square root. But which square root do we
want? Since the range of $f^{-1}$ is the domain of $f$, our answer
must be in $[0,\infty)$, so we want $f^{-1}(x) = y=+
\sqrt{(x-5)/2}$. This makes sense whenever $x \ge 5$. The domain of
$f^{-1}$ is $[5,\infty)$, which is the range of $f$.
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