Functions - f(x)

Concepts

$f (x) = x + 1$ , can be treated as $y = x + 1$

$f g (x) = f (g (x))$
So to get the expression for $f g (x)$ ,

Order matters

$f g (x) \neq g f (x)$

Alt syntax

$f g (x) = (f \circ g) (x)$
---------------^ circle

Let there be 2 functions:

Compositing the functions:

\begin{aligned} f g (x) & = f (\frac{x - 1}{x + 5}) \\ = {(\frac{x - 1}{x + 5})}^{2} - 1 \end{aligned}

\begin{aligned} f h (x) & = f (\sqrt{x + 1}) \\ = (\sqrt{x + 1})^{2} - 1 \\ = x + 1 - 1 \\ = x \end{aligned}

NOTE: Since $f h (x)$ gives x, it means that $f (x)$ and $g (x)$ are Inverse Function Example.

Syntax: $f^{- 1} (x)$
To solve,

Let $f (x) = x^{2} - 1$ , find the expression for $f^{- 1} (x)$

\begin{aligned} f (x) & = x^{2} - 1 \\ y & = x^{2} - 1 \end{aligned}

Swap x and y

\begin{aligned} x & = y^{2} - 1 \\ y^{2} & = x + 1 \\ y & = \sqrt{x + 1} \\ f^{- 1} (x) & = \sqrt{x + 1} \end{aligned}

You can composite the functions to shift, scale and reflect
For the examples below, let $f (x) = x^{2} + c$

	Vertically	Horizontally	Comments
Shift	$g (x) = f (x) + c$	$g (x) = f (x + c)$
Stretch & Compression	$g (x) = c f (x)$	$g (x) = f (c x)$	If $c < 0$ , it'll flip If $0 < c < 1$ , it'll compress

Transforming horizontally is opposite of what's expected

$g (x) = f (x - 3)$ will shift the graph up by 3 units
For example,

This also happens with stretching.
$g (x) = f (2 x)$ will compress the graph by half

Try to modify

c for shifting
d for stretching

Stretching might be hard to notice but it's scaling the graph horizontally/vertically depending if it's multiplied to the parameter/result