First, a problem:
$f(x) = |x|$
$g(x) = ln(x)$
$g \circ f(x) = ln(|x|)$
find domain($ g \circ f(x) $)
the function, $g \circ f(x)$ is well defined for $|x| > 0$, so for $x \neq 0$
$ \therefore \text{domain} (g \circ f) = \lbrace x \in \mathbb{R} | x \neq 0 \rbrace = (- \infty, 0) \cup (0, \infty)$
We restrict the natural domain of $f$ which is $\mathbb{R}$ to the domain $ X = (- \infty, 0) \cup (0, \infty)$
Then the composition $ x \rightarrow \ y \rightarrow z $ is well defined because range$(f) = (0, \infty)$ < domain($g$)
New question:
$ g \circ f = f \circ g$, True or False?
$ g \circ f = \ln{(|x|)}$,
$ f \circ g = f(g(x))$,
$ f \circ g = f( \ln{(x)} )$,
$ f \circ g = | \ln{(x)} |$,
$\therefore g \circ f \neq f \circ g$
Note: $\ln{(|x|)}$ is well defined for x = -1 but not for $| \ln{(x)}|$
domain($f \circ g$) = $ (0, \infty) $
domain($g \circ f$) = $ (-\infty, 0) \cup (0, \infty) $
$ g \circ f = \left(\begin{array}{rrrrr} \ln{(x)} \text{ if } x > 0 \\ \ln{(-x)} \text{ if } x < 0 \end{array}\right) $
$ f \circ g = \left(\begin{array}{rrrrr} \ln{(x)} \text{ if } \ln{(x)} > 0 \\ -\ln{(x)} \text{ if } \ln{(x)} < 0 \end{array}\right) $
$ \begin{array}{rrrrrrrr} \text{Sum } & (f + g)(x) = f(x) + g(x) \\ \text{Sub } & (f - g)(x) = f(x) - g(x) \\ \text{Constant Multiplication } & c \in \mathbb{R}, (c.f)(x) = c \times f(x) \\ \text{Product } & (f \times g)(x) = f(x) \cdot g(x) \\ \text{Quotient } & (f / g)(x) = f(x) / g(x) \\\ \text{Composition } & (f \circ g)(x) = g(f(x)) \end{array} $
Power functions: $x^n ,n \in \mathbb{Q}$, for example: $x^2, x^{-2}, x^{1/2}$
Trig functions: $ \sin, \cos, \tan $
$ \begin{array}{rrrrrrrr} \text{domain} (\sin{x}) = & \mathbb{R} \\ \text{domain} (\cos{x}) = & \mathbb{R} \\ \text{domain} (\tan{x}) = & \lbrace x \in \mathbb{R} | x \neq \frac{k \pi}{2}, k \in \mathbb{Z} \\ \text{range} (\sin{x}) = & [-1, 1] \\\ \text{range} (\cos{x}) = & [-1, 1] \\\ \text{range} (\tan{x}) = & \mathbb{R} \\\ \end{array} $