# How do you solve rational inequalities?

Oct 31, 2014

Let us solve the following rational inequality.

$f \left(x\right) = \frac{x + 1}{{x}^{2} + x - 6} \le 0$

Set the numerator equal to zero, and solve for $x$.

$x + 1 = 0 \implies x = - 1$

(Note: $f \left(- 1\right) = 0$)

Set the denominator equal to zero, and solve for $x$.

${x}^{2} + x - 6 = \left(x + 3\right) \left(x - 2\right) = 0 \implies x = - 3 , 2$

(Note: $f \left(- 3\right)$ and $f \left(2\right)$ are undefined.)

Using $x = - 3 , - 1 , 2$ above to split the number line into open intervals:

$\left(- \infty , - 3\right) , \left(- 3 , - 1\right) , \left(- 1 , 2\right)$, and $\left(2 , \infty\right)$

Using sample numbers $x = - 4 , - 2 , 0 , 3$ for each interval above, respectively, we can determine the sign of (LHS).

$f \left(- 4\right) = - 2 < 0 \implies f \left(x\right) < 0$ on $\left(- \infty , - 3\right)$

$f \left(- 2\right) = \frac{1}{4} > 0 \implies f \left(x\right) > 0$ on $\left(- 3 , - 1\right)$

$f \left(0\right) = - \frac{1}{6} < 0 \implies f \left(x\right) < 0$ on $\left(- 1 , 2\right)$

$f \left(3\right) = \frac{2}{3} > 0 \implies f \left(x\right) > 0$ on $\left(2 , \infty\right)$

Hence, $f \left(x\right) \le 0$ on $\left(- \infty , - 3\right) \cup \left[- 1 , 2\right)$.

(Note: $- 1$ is included since $f \left(- 1\right) = 0$.)

The graph of $y = f \left(x\right)$ looks like:

I hope that this was helpful.