2
$\begingroup$

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.

$\endgroup$

1 Answer 1

6
$\begingroup$

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 $r$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.