Inverse of Function

Just flip the x and y values

y=x^{2}
y=x
x=y^{2}

reflection along y=x
Domain and ranges are just flipped
Inversion may not always be a function
- If it is, it is notated as $f^{- 1} (x)$

How to find inverse of function

$\begin{aligned} f (x) & = - 2 \frac{1}{x + 2} - 3 \\ D_{f} & : {x | x \neq - 2, x \in R} \\ R_{f} & : {y | y \neq - 3, y \in R} \\ y & = - 2 \frac{1}{x + 2} - 3 \\ For the inverse \\ x & = - 2 \frac{1}{y + 2} - 3 \\ x + 3 & = - 2 \frac{1}{y + 2} \\ \frac{1}{- 2} (x + 3) & = \frac{1}{y + 2} \\ y + 2 & = \frac{1}{- \frac{1}{2} (x + 3)} \\ y & = \frac{1}{- \frac{1}{2} (x + 3)} - 2 \\ D_{f^{- 1}} & : {x | x \neq - 3, x \in R} \\ R_{f^{- 1}} & : {y | y \neq - 2, y \in R} \end{aligned}$

y=-2 \frac{1}{x+2}-3
y=\frac{1}{-\frac{1}{2}(x+3)}-2
y=x

If $f (x) = f^{- 1} (x)$ , it is it's self-inverse
- That is true if the graph is symmetrical along $y = x$
- eg $y = \frac{1}{x}$
formalized
$y = f^{- 1} (x) \Leftrightarrow x = f (y)$
Domain of $f^{- 1}, D (f^{- 1})$ is the range of $f, R (f)$
$R (f^{- 1}) = D (f)$
$f^{- 1} (f (x)) = x$ for all $x \in D (f)$
$f (f^{- 1} (x)) = x$ for all $x \in D (f^{- 1})$
$(f^{- 1})^{- 1} (x) = f (x)$ for all $x \in D (f)$