Single Variable Calculus, Chapter 7, 7.1, Section 7.1, Problem 30

Determine the explicit function for $f^{-1}$ if $f(x) = \sqrt{x^2 + 2x}$ for $x > 0$ and use it to graph $f^{-1}$, f and the line $y = x$ on the same
If $f(x) = \sqrt{x^2 + 2x}$, then

$
\begin{equation}
\begin{aligned}
f^{-1} (x) \quad \Longrightarrow \quad x &= \sqrt{y^2 + 2y}\\
\\
x^2 &= y^2 + 2y\\
\\
\text{by completing the square}\\
\\
x^2 + 1 &= y^2 + 2y + 1\\
\\
x^2 + 1 &= (y+1)^2\\
\\
\pm \sqrt{x^2+1} &= y + 1\\
\\
y &= -1 \pm \sqrt{x^2 + 1}
\end{aligned}
\end{equation}
$

We got the two values of $y$, however, the function is defined only for $x > 0$. The domain of $f$ is $x > 0$ and its range $y > \sqrt{0^2 + 2(0)} \Longrightarrow 0$. Thus, the domain of $f^{-1}(x)$ is $x > 0$ and range $y > 0$. Thus, the domain of $f^{-1}(x)$ is $x > 0$ and range $y > 0$. If we check both $y$,
when $ x = 1$,

$
\begin{equation}
\begin{aligned}
y &= -1 + \sqrt{1^2 + 1}\\
\\
y &= 0.4142
\end{aligned}
\end{equation}
$


when $ x = 1$,

$
\begin{equation}
\begin{aligned}
y &= -1 - \sqrt{1^2 + 1}\\
\\
y &= -2.4142
\end{aligned}
\end{equation}
$

Hence, $f^{-1}(x) = -1 + \sqrt{x^2+1}$