Existence - Composite functions

Motivating example

Consider the functions

\begin{align*} &f: x \mapsto \frac{1}{x-3}, &&\textrm{for } x \in \mathbb{R}, x \neq 3 \\ &g: x \mapsto x+2, &&\textrm{for } x \in \mathbb{R}. \end{align*}

In trying to set up the composite function ${fg}$ , we run into a problem if we try to substitute ${x=1}$ .

The first step in getting ${g(1)=3}$ works as intended but trying to substitute ${3}$ into ${f}$ does not work as ${3}$ is not within the domain of ${f}$ .

This example gives us an insight into the condition for the existence of the composite function ${fg}$ : that the range of ${g}$ must be a subset of the domain of ${f}$ .

The composite function $fg$ exists if $R_g \subseteq D_f$ .

Visualization

Examples

Consider the functions

egin{align*} &f: x mapsto rac{1}{x-3}, && extrm{for } x in mathbb{R}, x eq 3 \ &g: x mapsto x+2, && extrm{for } x in mathbb{R}. end{align*}

{R_g = (-\infty, \infty)}

egin{align*} &f: x mapsto rac{1}{x-3}, && extrm{for } x in mathbb{R}, x eq 3 \ &g: x mapsto x+2, && extrm{for } x in mathbb{R}. end{align*}

{R_f = (-\infty, 0) \cup (0, \infty)}