Paul's Online Notes
Paul's Online Notes
Home / Algebra / Solving Equations and Inequalities / Rational Inequalities
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 2.13 : Rational Inequalities

5. Solve the following inequality.

\[u \le \frac{4}{{u - 3}}\]

Show All Steps Hide All Steps

Start Solution

The first thing we need to do is get a zero on one side of the inequality and then, if possible, factor the numerator and denominator as much as possible.

So, we first need to get zero on one side of the inequality.

\[u - \frac{4}{{u - 3}} \le 0\]

We now need to combine the two terms in to a single rational expression.

\[\begin{align*}\frac{{u\left( {u - 3} \right)}}{{u - 3}} - \frac{4}{{u - 3}} & \le 0\\ \frac{{{u^2} - 3u - 4}}{{u - 3}} & \le 0\end{align*}\]

Finally, we need to factor the numerator.

\[\frac{{\left( {u - 4} \right)\left( {u + 1} \right)}}{{u - 3}} \le 0\]
Hint : Where are the only places where the rational expression might change signs?
Show Step 2

Recall from the discussion in the notes for this section that the rational expression can only change sign where the numerator is zero and/or where the denominator is zero.

We can see that the numerator will be zero at,

\[u = - 1\hspace{0.25in}\hspace{0.25in}u = 4\]

and the denominator will be zero at,

\[u = 3\]
Hint : Knowing that the rational expression can only change sign at the points above how can we quickly determine if the rational expression is positive or negative in the ranges between those points?
Show Step 3

Just as we did with polynomial inequalities all we need to do is check the rational expression at test points in each region between the points from the previous step. The rational expression will have the same sign as the sign at the test point since it can only change sign at those points.

Here is a sketch of a number line with the points from the previous step graphed on it. We’ll also show the test point computations on the number line as well. Here is the number line.

Show Step 4

All we need to do now is get the solution from the number line in the previous step. Here is both the inequality and interval notation from of the answer.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{array}{c} u \le - 1\hspace{0.25in}{\mbox{and }}\hspace{0.25in} 3 < u \le 4\\ \left( { - \infty , - 1} \right] \hspace{0.25in} {\mbox{and }}\hspace{0.25in} \left( {3,4} \right]\end{array}}\]

Be careful with the endpoints for this problem. Because we have an equal sign in the original inequality we need to include \(u = - 1\) and \(u = 4\) because the numerator and hence the rational expression will be zero there. However, we can’t include \(u = 3\) because the denominator is zero there and so the rational expression has division by zero at that point!