College Algebra, Chapter 3, 3.7, Section 3.7, Problem 24

If $g(x) = x^2 + 4x$ with $x \geq -2$, find $g^{-1} (5)$.

To find the inverse of $g(x)$, write $y = g(x)$


$
\begin{equation}
\begin{aligned}

y =& x^2 + 4x
&& \text{Model}
\\
\\
y + 4 =& x^2 + 4x + 4
&& \text{If we solve for $x$, we use completing the square: add } \left(\frac{4}{2} \right)^2 = 4
\\
\\
y + 4 =& (x + 2)^2
&& \text{Perfect Square}
\\
\\
\pm \sqrt{y + 4} =& x + 2
&& \text{Take the square root}
\\
\\
x =& -2 \pm \sqrt{y + 4}
&& \text{Subtract } 2
\\
\\
y =& -2 \pm \sqrt{x + 4}
&& \text{Interchange $x$ and $y$}
\\
\\
\text{Thus,} &
&&
\\
\\
g^{-1} (x) =& -2 \pm \sqrt{x + 4}
&&
\\
\\
\text{Thus, } g^{-1} (5) =& -2 + \sqrt{5 + 4} \qquad \text{ and }
&& g^{-1} (5) = -2 - \sqrt{5 + 4}
\\
\\
=& -2 + 3
&&= -2 - 3
\\
\\
=& 1
&&= -5



\end{aligned}
\end{equation}
$


Last, we have restrictions $x \geq -2$, so..

$g^{-1} (5) = 1$