# What is the domain and range of the rational function $\frac{x-5}{2x+1}$?

This is a homework question from a precalculus class that I'm a TA for.

What is the domain and range of the function $$f(x) = \frac{x-5}{2x+1}\,?$$

I wanted to write up a thorough solution to this exercise for my class, and figured I'd post it online to help anyone else who may wander across it.

Since $$f$$ is a rational function, the only restriction that we need to impose on the domain of $$f$$ will be to exclude values of $$x$$ for which the denominator of $$f$$ is zero$$\ast$$, since dividing by zero is an undefined operation. The denominator equals zero when $$x = -\frac{1}{2}$$, so the domain of $$f$$ will consist of all real number except for $$-\frac{1}{2}$$. If we felt inclined to, we could express this domain as either of the following: $$\left(-\infty,-\frac{1}{2}\right) \cup \left(-\frac{1}{2},\infty\right)\qquad\qquad \mathbb{R}\setminus\left\{-\frac{1}{2}\right\}$$
Now for the range of $$f$$, we need to find the set of all $$y$$ for which there exists some $$x$$ where $$y = \frac{x-5}{2x+1}\,.$$ We can rearrange this equation, solving for $$x$$ in terms of $$y$$, to give us an explicit formula for which $$x$$ will result in a given $$y$$. Then we can determine the range of $$f$$ by noting for which $$y$$ the formula will be undefined. \begin{align*} y &= \frac{x-5}{2x+1} \\[0.7em] (2x+1)y &= x-5 \\[0.7em] 2xy-x &= -5-y \\[0.7em] x &= -\frac{5+y}{2y-1} \end{align*} This formula is defined for all $$y \neq \frac{1}{2}$$; for every other real number $$y$$ we can use this formula to find a corresponding $$x$$ such that $$f(x)=y$$. So $$\frac{1}{2}$$ is the only value which is not in the range of $$f$$, and we can express the range of $$y$$ as either of the following:
$$\left(-\infty,\frac{1}{2}\right) \cup \left(\frac{1}{2},\infty\right)\qquad\qquad \mathbb{R}\setminus\left\{\frac{1}{2}\right\}$$
$$\ast$$ Thoroughly justifying this statement is kinda complicated. I wouldn't expect my precalculus class to justify this. I suppose if you really wanted to though, you could proceed by showing $$f$$ can be written as a composite of the functions $$r(x) = \frac{1}{x} \qquad S_a(x) = ax \qquad T_b(x) = x+b \qquad\text{for } a,b \in \mathbb{R}\,,$$ and since $$S_a$$ and $$T_b$$ are bijective functions $$\mathbb{R} \to \mathbb{R}$$, so the only domain restrictions we have to impose on $$f$$ are those that come from the function $$r$$.