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 $\displaystyle f^{-1} (x) = \sqrt{x + \frac{1}{4}}$.