Calculus: Early Transcendentals, Chapter 1, Review Exercises, Section Review Exercises, Problem 24

Determine the inverse function of $\displaystyle f(x) = \frac{x + 1}{2x + 1}$

We first write $\displaystyle y = \frac{x + 1}{2x + 1}$

Then we solve this equation for $x$

$
\begin{equation}
\begin{aligned}
y(2x + 1) &= x + 1\\
\\
2xy + y &= x + 1\\
\\
2xy - x &= 1 - y \\
\\
x(2y - 1) &= 1 -y \\
\\
x &= \frac{1 - y}{2y - 1}
\end{aligned}
\end{equation}
$


Finally, we interchange $x$ and $y$
$\displaystyle y = \frac{1 - x}{2x - 1}$
Therefore, the inverse fraction is
$\displaystyle f^{-1}(x) = \frac{1 - x}{2x - 1}$