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?

  • $\begingroup$ First simplify the expression then compute domain and codomain. $\endgroup$
    – Matteo
    Jul 31, 2019 at 6:02
  • 2
    $\begingroup$ @Matteo That is not the right approach. See my answer. $\endgroup$
    – Anurag A
    Jul 31, 2019 at 6:28

1 Answer 1


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 $\color{blue}{\Bbb{R}-\{1,\frac{5}{9}\}}.$


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