Given the composition of function $f(g(x))$ and $g(x)$, what is the domain of $f(x)$?

Question

Suppose that I have $g(x)=\frac{1}{x}$ we know that the domain is all real numbers except 0 and $f(g(x))=\frac{2}{x}+5$. we know the domain is also all real numbers except 0 We know that to find $f(x)$, what we have to do is find the inverse function of $g(x)$, which is $g^{-1}(x)=\frac{1}{x}$. Then we substitute it and get $f(g(g^{-1}(x)))=2x+5$, so that $f(x)=2x+5$.

Does it affect the domain of $f(x)$? I mean, when you substitute $x$ with $g^{-1}(x)$ to $f(g(x))$, then the domain of $f(x)$ must be the domain of $g^{-1}(x)$, right (that is the domain of $f(x)$ is $x$ not $0$ which is same as the domain of $g^{-1}(x)$)? Please correct me if I am wrong.

Answer 1

Usually, the domain of a function is part of its definition and cannot be obtained from its formula. You can ask questions like: What is the largest/smallest set on which $f$ can be defined s.t. it fulfills the requirements? But asking for the domain of a function does not have a mathematical meaning.

Having said that, in school there are often questions of that type, which are supposed to ask for the largest set on which some expression is defined. This is, in your case, trivial: $f(x)=2x+5$ can clearly be defined everywhere.

A slightly more interesting question, and what you probably intended to ask, is: What is the smallest possible domain on which $f$ has to be defined so that $f(g(x))=\frac2x+5$ holds for every $x\in\operatorname{dom}(g)=\mathbb R\setminus\{0\}$? Here, $\operatorname{dom}$ stands for the domain of a function and the statement $\operatorname{dom}(g)=\mathbb R\setminus\{0\}$ should be part of the definition of $g$. For $f(g(x))$ to be defined, $f$ must be defined for every $x$ in the image of $g$, that is it has to hold that$$\operatorname{im}(g):=\{g(x)\mid x\in\operatorname{dom}(g)\}\subseteq\operatorname{dom}(f).$$ To find the image of $g$, you can use the inverse function: One finds $$\operatorname{im}(g)=\operatorname{dom}(g^{-1})\overset{g^{-1}=g}=\operatorname{dom}(g)=\mathbb R\setminus\{0\}.$$ This means that $f$ has to befined everywhere except at $0$ for your conditions to hold for all $x\in\operatorname{dom}(g)$. (But could of course also be defined at $0$.)

Answer 2

Let's frame it as follows: Define $g,h:(0,\infty)\to(0,\infty)$ by $g(x):=\frac{1}{x}$ and $h(x):=\frac{2}{x}+5$. We want to investigate the possible domains of a function $f$ satisfying $fg=h$.

Evaluating $fg=h$ at $x>0$ gives that $f(\frac{1}{x})=2(\frac{1}{x})+5$, so that $f(y)=2y+5$ for all $y>0$. Now let $(0,\infty)\subseteq A\subseteq\mathbb{R}$ and define $f:A\to\mathbb{R}$ by $f(x):=2x+5$.

Then $(fg)(x)=h(x)$ for all $x>0$. So this works for any domain containing $(0,\infty)$. In fact, it doesn't matter what values $f$ takes for $x\leq 0$, since $g$ maps only into $(0,\infty)$.

But this depends on the domain of $g$ and $h$, which are defined for all non-zero $x$. In short, there is no definitive answer to your question as stated. The domain of a function is almost always included as part of the definition. You might be able to work out where it needs to be defined for some particular purpose, but very often you could tack on some extra elements to the domain, and assign arbitrary values to those extra elements, without affecting anything all that much.