Existence of composite functions

Introduction to composite functions

A composite function applies functions in sequence.

For example, consider the functions

$\begin{aligned} f : x \mapsto 2 x, & x \in ℝ \\ g : x \mapsto x + 1, & x \in ℝ . \end{aligned}$

Starting from $x = 2,$ we could first apply the function $f$ to get $f (2) = 4 .$ We then apply the function $g$ to get $g (f (2)) = g (4) = 5 .$

The composite function $g f$ represents the overall function that gets us from the starting value of $x = 2$ to the final value of $x = 5 .$

Note that while we read left to right in English, the composite function $g f$ applies $f$ first followed by $g .$

Order of composition is important. The function $f g$ applies $g$ first followed by $f$ and is a different function. For our example, $f g (2) = 6 .$

Composite functions may not exist

Now consider the following functions:

$\begin{aligned} f : x \mapsto \frac{1}{x}, & x \in ℝ, x \neq 0 \\ g : x \mapsto x + 1, & x \in ℝ . \end{aligned}$

The function $f g$ does not exist because if we were to start with $x = - 1,$ we get $g (- 1) = 0$ and will be unable to continue applying $f$ as zero is not within the domain of $f .$

Criteria for composite to exist

To prevent situations like the example above, all the outputs of the first function in a composite function must lie within the domain of the second function. That is, for $f g$ to exist, the range of $g$ must be a subset of the domain of $f .$

The composite function $f g$ exists if $R_{g} \subseteq D_{f} .$
The composite function $f g$ does not exist if $R_{g} \subseteq̸ D_{f} .$

Example

Question:

$\begin{aligned} f : x \mapsto 2 x - 1, & 0 \leq x < 5 \\ g : x \mapsto x^{2}, & - 2 < x < 1 . \end{aligned}$

Determine if the composite functions $g f$ and $f g$ exist.

Solution:
$R_{f} = [- 1, 9)$ and $D_{g} = (- 2, 1) .$ Hence $R_{f} \subseteq̸ D_{g}$ so the composite function $g f$ does not exist. $∎$

$R_{g} = [0, 4)$ and $D_{f} = [0,5) .$ Hence $R_{g} \subseteq D_{f}$ so the composite function $f g$ exists. $∎$