Rational Inequalities

We can use the same steps to solve rational inequalities graphically as we did to solve polynomial inequalities graphically.

Steps for Solving Rational Inequalities Graphically

1. Rewrite the equation or inequality, if necessary, so that an expression is on one side with 0 on the other side.

2. Set the expression of the equation or inequality equal to ƒ(x), and graph the related function.

3. Use the graph of ƒ(x) to determine solutions as follows.

   1. The real solutions of ƒ(x) = 0 are the x-values of the
      x-intercepts of the graph. These are the zeros of ƒ(x).

   2. The real solutions of ƒ(x) < 0 are the x-values for which the graph lies below the x-axis.

   3. The real solutions of ƒ(x) > 0 are the x-values for which the graph lies above the x-axis.

Example

Solve $\frac{x-5}{x+2}\ge0.$

Solution

The inequality is already written with 0 on one side, so we are ready to graph.

The vertical asymptote has equation x = -2, and the horizontal asymptote has equation y = 1. The x-intercept, found by setting the numerator equal to 0, is (5, 0).

Evaluating ƒ(0) gives the y-intercept (0, -5/2). The graph does not intersect its horizontal asymptote because ƒ(x) = 1 has no solution.

$1=\frac{x-5}{x+2}$

$x+2=x-5$

$2=-5$

The solution set includes the x-values for which the graph of f(x) lies on or above the x-axis. Because the inequality is nonstrict, the zero of f(x),that is x = 5, is included in the solution set. The solution set is ($-\infty,-2$)$\cup$[5,$\infty$).

Example

Solve $\frac{2}{x+3}<\frac{1}{x-1}$

Solution

$\frac{2}{x+3}-\frac{1}{x-1}<0$

$\frac{2\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{1\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}<0$

$\frac{2\left(x-1\right)-\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}<0$

$\frac{2x-2-x-3}{\left(x+3\right)\left(x-1\right)}<0$

$\frac{x-5}{\left(x+3\right)\left(x-1\right)}<0$

The vertical asymptotes have equations x = -3 and x = 1. The horizontal asymptote has equation y = 0. The y-intercept is (0, 5/3), and the x-intercept is (5, 0) The graph intersects its horizontal asymptote at (5, 0). Additional points may be used as necessary to sketch the graph.

As f(x) < 0, the graph lies below the x-axis for solution set $\left(-\infty,-3\right)\cup\left(1,5\right).$Because the inequality is strict, the zero of f(x) at x=5 is not included in the set.

Examples

Here is a some video with some more examples: