If $f(x)=1/x$, and $g(x)=x+1$, is zero in the domain of $f(g(x))$?

Question

I’m a little confused on domain restrictions in compositions of functions. $f(g(x))$ equals $1/(x+1)$, so $-1$ is not in the domain obviously, but what about zero? $f$ originally excluded $0$ from the domain, so is it still excluded in the composition?

Answer 1

Any value where $g(x)$ or $f(g(x))$ is undefined will be excluded from the domain, but not necessarily $f(x)$. $g(x) = x+1$ and $f(g(x)) = \frac{1}{x+1}$. There are no numbers in $\mathbb{R}$ where $g(x)$ is undefined, and $-1$ is undefined for $f(g(x))$. Thus, the domain is $(-\infty, -1)\cup (-1, \infty)$.

Answer 2

What is excluded from the domain is anything that gives you an invalid input to any function. For these two functions, the only invalid input is to give the input $0$ to $f$. In other words, the only invalid input to the composite function is $x=-1$.

Answer 3

As you say $f(g(x))=\frac{1}{x+1}$. So the only value not in the domain is $x=-1$. Plugging in $0$ is fine.

$f$ by itself has domain $\mathbb R\setminus \{0\}$. That means when looking at the composite function $f(g(x))$, $g(x)$ is not allowed to be $0$.

In general, if $g:A\to B$ and $f:B\to C$, then the domain of $f\circ g$ is $g^{-1}(B)$. That just means that both $g(x)$ and $f(g(x))$ must be well-defined.