91Ó°ÊÓ

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

$\frac{6}{3x^2−2x−5}â‰\yen 0$

The given function is$\frac{6}{3x^2−2x−5}â‰\yen 0$

The inequality is in the correct form.

Factor the denominator.

In the equation the product of first and last term is $15$and second term is $2$. So the possible factors are $3$and $5$.

$$\begin{gathered} =3x^2−(5-3)x−5 \\ =3x^2−5x+3x−5 \\ =x(3x-5)+1(3x-5) \\ =(x+1)(3x-5) \end{gathered}$$

$\frac{6}{(x+1)(3x-5)}â‰\yen 0$

The quotient is 0 when the numerator is 0.

Since the numerator is always 6, the quotient cannot be 0.

The quotient will be undefined when the denominator is zero.

$$\begin{gathered} x+1=03x-5=0 \\ x=-1x=\frac{5}{3} \end{gathered}$$

The number line is divided into three intervals:$(-∞,-1)(-1,\frac{5}{3})(\frac{5}{3},∞)$

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

$\frac{6}{(x+1)(3x-5)}=\frac{6}{(-2+1)\left(3\right(-2)-5)}=\frac{6}{-1·(-6-5)}=\frac{6}{11}$

The quotient is marked positive in the interval $(-∞,-1)$.

The number 1 is in the interval $(-1,\frac{5}{3})$. Test $x=1$in the expression in the numerator and the denominator.

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

The quotient is marked negative in the interval $(-1,\frac{5}{3})$.

The number $3$is in the interval $(\frac{5}{3},∞)$. Test $x=3$in the expression in the numerator and the denominator.

$\frac{6}{(x+1)(3x-5)}=\frac{6}{(3+1)(3×3-5)}=\frac{6}{4(9-5)}=\frac{6}{4·4}=\frac{3}{8}$

The quotient is marked positive in the interval $(\frac{5}{3},∞)$.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals$(-∞,-1)$and$(\frac{5}{3},∞)$are solutions.

The solution in interval notation is $(-∞,-1)∪(\frac{5}{3},∞)$.