# Domain and Codomain of Composite Functions

I'm wondering if I'm over complicating my understanding of domain and codomain for composite functions.

Say, for example, I have $$f(x)=\frac{x}{x+2}$$, then I know the domain is $$x \neq -2, x \in \Re$$, and the codomain is $$y \neq 1, y \in \Re$$.

Then I could also have $$g(x)=\frac{5x}{2x-3}$$, then I know the domain is $$x \neq \frac{3}{2}, x \in \Re$$, and the codomain is $$y \neq \frac{5}{2}, y \in \Re$$.

If I have a composition of $$f \circ g$$, then I know that I have $$f(g(x))=\frac{\frac{5x}{2x-3}}{\frac{5x}{2x-3}+2}$$, but in terms of domain and codomain, do I consider the original functions, or do I only consider the new composition?

My original idea was that the domain and codomain of $$f \circ g$$ would not only reject the values of the domains of $$f$$ and $$g$$, but also reject anything that would eventually not be computed.

I'd have $$x \neq \frac{3}{2}, -2, x \in \Re$$ because of the separate functions themselves. But then I'd have to include what $$f$$ would have been if the restriction on $$g$$'s codomain would have been allowed, so without $$g(x)=\frac{5}{2}$$, I couldn't get $$f(\frac{5}{2})=\frac{5}{9}$$

So I'd actually say that the domain of $$f \circ g$$ is $$x \neq \frac{3}{2},-2, \frac{5}{2}$$ with $$x \in \Re$$, and then the codomain is just $$y \neq \frac{5}{2},\frac{5}{9}, y \in \Re$$... but I think it's too many restrictions?

• First simplify the expression then compute domain and codomain. Jul 31, 2019 at 6:02
• @Matteo That is not the right approach. See my answer. Jul 31, 2019 at 6:28

When considering $$g: A \rightarrow B$$ and $$f: C \rightarrow D$$. For $$f \circ g$$ to be defined $$g(A) \subseteq C$$, otherwise we take that subset $$T \subset A$$ for which $$g(T) \subseteq C$$ and thus the domain of $$f \circ g$$ may not be exactly $$A$$.
With that in mind. In your example, $$g:\Bbb{R}-\{3/2\} \longrightarrow \Bbb{R}-\{5/2\}, \qquad f:\Bbb{R}-\{-2\} \longrightarrow \Bbb{R}-\{1\}.$$ So for $$f(g(x))$$ to be defined, we can choose any value for $$x \in \Bbb{R}-\{3/2\}$$ except the one for which $$g(x)=-2$$ (because $$f$$ cannot have $$-2$$ as an input). Thus $$x \neq \frac{2}{3}$$ as well.
Thus domain of $$f(g(x))$$ will be $$\color{red}{\Bbb{R}-\{\frac{3}{2}, \frac{2}{3}\}}$$.
For the range of the composition, bear in mind that the range of $$f \circ g$$ must be a subset of the range of $$f$$. So for sure $$\text{range}(f \circ g)$$ will not have $$1$$. Furthermore $$g$$ does not output the value $$5/2$$, this means we will not have $$f(5/2)=5/9$$ also in the range of the composition function.
Thus the range of $$f \circ g$$ is $$\color{blue}{\Bbb{R}-\{1,\frac{5}{9}\}}.$$