Self-inverse function

If $f$ is a self-inverse function, $$ \boxed{ f^2 (x) = f f (x) = x } $$

Example

Given $$ g : x \mapsto {x + 5 \over 2x - 1}, \phantom{000} x \in \mathbb{R}, x \ne {1 \over 2} $$

Forming the composite function $g^2$: \begin{align} g^2 (x) & = g \left( {x + 5 \over 2x - 1} \right) \\\ & = { {x + 5 \over 2x - 1} + 5 \over 2 \left({x + 5 \over 2x - 1} \right) - 1} \\ & = { {x + 5 \over 2x - 1} + 5 \over 2 \left({x + 5 \over 2x - 1} \right) - 1} \times {2x - 1 \over 2x - 1} \\ & = {x + 5 + 5(2x - 1) \over 2(x + 5) - (2x - 1) } \\ & = {x + 5 + 10x - 5 \over 2x + 10 - 2x + 1} \\ & = {11x \over 11} \\ & = x \end{align}

Thus, $g$ is a self-inverse function.

Inverse function of a self-inverse function

If $f$ is a self-inverse function, $ff (x) = x$

Since $ f^{-1} f (x) = x$, $$ \boxed{ f^{-1} (x) = f(x) } $$

Example

Given $$ g : x \mapsto {x + 5 \over 2x - 1}, \phantom{000} x \in \mathbb{R}, x \ne {1 \over 2} $$

Forming the inverse function: \begin{align} \text{Let } y & = {x + 5 \over 2x - 1} \\ y(2x - 1) & = x + 5 \\ 2yx - y & = x + 5 \\ 2yx - x & = y + 5 \\ x(2y - 1) & = y + 5 \\ x & = {y + 5 \over 2y - 1} \\ \\ g^{-1} (x) & = {x + 5 \over 2x - 1} \end{align}