Prohibited values when plugging one formula into another?

Question

Say f(x)=$\frac{1}{x}$

and g(x)=$x-5$

and we where to plug g into f as a composite function, would x = 0 still remain a prohibit value of f or not?

Thanks!

Answer 1

We have $f(g(x))=\frac{1}{x-5}$, hence $f \circ g$ is defined at$x=0$.

Answer 2

$f$ has $\mathbb{R}\setminus \{0\}$ as its domain and also as its range, while $g$ has $\mathbb{R}$ as its domain and range.

So the domain of $g \circ f$ is still $\mathbb{R}\setminus\{0\}$ but $f \circ g$ (so "plugging $g$ into $f$") has $\mathbb{R}\setminus \{5\}$ as its domain, because we must make sure that $f$ can be applied to the value of $g$, so any $x$ with $g(x) = 0$ is off, as $0$ is not in $f$'s domain. So $x=5$ is not in the domain of the composition (and all other values are).

Answer 3

In detail, \begin{align} f&:\mathbb R\setminus\{0\} \to \mathbb R \\&:x \mapsto \frac 1x \end{align} and \begin{align} g&:\mathbb R \to \mathbb R \\&:x \mapsto x-5 \end{align}

In order to compute the composition $f\circ g$ it must be true that the image of $g$ is a subset of the domain of $f$. This isn't true since $\mathbb R \not \subseteq \mathbb R\setminus\{0\}$.

This is a really fancy way of saying that, since we can't compute $f(g(5))$, then we can't make $f \circ g$.

It is very common though to say that $f(g(x)) = \dfrac{1}{x-5}$ except when $x = 5$. To do this formally, however, you would have to redefine $g$ as

\begin{array}{llclcl} g &: &\mathbb R \setminus \{5\} &\to &\mathbb R \setminus \{0\} \\ &: &x &\mapsto &x-5 \end{array}