Question on domain and ranges.

Question

Suppose that $f(x) = 13x-4$ and $g(x) = \frac{1}{6x-18}$. Then clearly \begin{align*} \text{Dom}f &=(-\infty, \infty)\\ \text{Ran}f &= (-\infty , \infty )\\ \text{Dom}g &= (-\infty, 3)\cup (3, \infty)\\ \text{Ran}g &= (-\infty, 0)\cup (0, \infty ) \end{align*} Now considering $(f\circ g)(x)$ we have $$(f\circ g)(x) = 13\left( \frac{1}{6x-18} \right) -4 = \frac{13}{6x-18}-4\frac{6x-18}{6x-18} = \frac{13-24x+72}{6x-18}=\frac{85-24x}{6x-18}$$ so that \begin{align*} \text{Dom}\left( f\circ g \right) &= (-\infty, 3)\cup (3, \infty)\\ \text{Ran}\left( f\circ g \right) &=(-\infty, -4)\cup (-4,0)\cup (0, \infty) \end{align*} My question is regarding the range. I am aware that if the composition of the functions punctures a new hole in the domain then I have to exclude it from the domain of the composition. But my question is: does the same follow for the range? That is, should the range of the composition above really be $(-\infty, -4) \cup (-4, \infty )$ or is it correct as is? If so why?

Answer 1

The range of $f\circ g$ is $\Bbb R\setminus\{-4\}$. The number $0$ belongs to the range, since$$f\left(g\left(\frac{85}{24}\right)\right)=0.$$