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..

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