Self-inverse Function
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H2 Maths Formulas, Techniques & Graphs >> Functions and Graphs >> Functions >>
Self-inverse functionA self-inverse function ‘reverses itself’ to produce the original input:
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}
More on Functions:
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Find range of function by GC
Inverse function
Test for 1-1 function - Horizontal line test
Rule, domain, range & graph of inverse functions
Composite functions
Conditions for composite function to exist
Find the range of composite function
Other concepts (on Functions)
Composite functions f⁻¹f and ff⁻¹
Other formulas, techniques & graphs
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