In the following exercises, solve each rational inequality and write the solution in interval notation.

$\frac{1}{x^2-4x-12}>0$

In the following exercises, we have to solve rational inequality and write the solution in interval notation.

$\frac{1}{x^2-4x-12}>0$

The given function is$\frac{1}{x^2-4x-12}>0$

The inequality is in the correct form.

Factor the denominator.

In the equation the product of first and last term is $12$and the second term is $4$. So the possible factors are $2$and $6$.

$$\begin{gathered} =x^2-(6-2)x-12 \\ =x^2-6x+2x-12 \\ =x(x-6)+2(x-6) \\ =(x+2)(x-6) \end{gathered}$$

$\frac{1}{(x+2)(x-6)}>0$

The quotient is 0 when the numerator is 0.

Since the numerator is always , the quotient cannot be 0.

The quotient will be undefined when the denominator is zero.

$$\begin{gathered} x+2=0x-6=0 \\ x=-2x=6 \end{gathered}$$

The number line is divided into three intervals:

$(-\infty ,-2)(-2,6)(6,\infty )$

The number $-3$is in the interval $(-\infty ,-2)$. Test $x=-3$in the expression in the numerator and the denominator.

$\frac{1}{(x+2)(x-6)}=\frac{1}{(-3+2)(-3-6)}=\frac{1}{(-1)·(-9)}=\frac{1}{9}$

The quotient is marked positive in the interval $(-\infty ,-2)$.

The number $1$is in the interval $(-2,6)$. Test $x=1$in the expression in the numerator and the denominator.

$\frac{1}{(x+2)(x-6)}=\frac{1}{(1+2)(1-6)}=\frac{1}{3·(-5)}=\frac{1}{-15}$

The quotient is marked negative in the interval $(-2,6)$.

The number $7$is in the interval $(6,\infty )$. Test $x=7$in the expression in the numerator and the denominator.

$\frac{1}{(x+2)(x-6)}=\frac{1}{(7+2)(7-6)}=\frac{1}{9·1}=\frac{1}{9}$

The quotient is marked positive in the interval $(6,\infty )$.

We want the quotient to be greater than zero, so the numbers in the intervals $(-\infty ,-2)$and $(6,\infty )$are solutions.

The solution in interval notation is $(-\infty ,-2)\cup (6,\infty )$.