College Algebra, Chapter 3, 3.7, Section 3.7, Problem 46

Find the inverse of $\displaystyle f(x) = x^2 + x; x \geq - \frac{1}{2}$

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


$\begin{equation} \begin{aligned} y =& x^2 + x && \text{Solve for $x$ by completing the square: add } \left( \frac{1}{2} \right)^2 = \frac{1}{4} \\ \\ y + \frac{1}{4} =& x^2 + x + \frac{1}{4} && \text{Perfect Square} \\ \\ y + \frac{1}{4} =& \left( x + \frac{1}{2} \right)^2 && \text{Take the square root} \\ \\ \pm \sqrt{y + \frac{1}{4}} =& x + \frac{1}{2} && \text{Subtract } \frac{1}{2} \\ \\ x =& \frac{-1}{2} \pm \sqrt{y + \frac{1}{4}} && \text{Interchange $x$ and $y$} \\ \\ y =& \frac{-1}{2} \pm \sqrt{x + \frac{1}{4}} && \text{Apply restrictions } x \geq \frac{-1}{2} \\ \\ y =& \frac{-1}{2} + \sqrt{x + \frac{1}{4}} && \end{aligned} \end{equation}$


Thus, the inverse of $f(x) = x^2 + x$ is $\displaystyle f^{-1} (x) = \sqrt{x + \frac{1}{4}}$.