Question

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

$\frac{-2}{6x^2-13x+6}\le 0$

Short answer

The solution in interval notation is$(-\infty ,\frac{2}{3})\cup (\frac{3}{2},\infty )$

Step-by-step solution

Step 1. Given Information  

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

$\frac{-2}{6x^2-13x+6}\le 0$

Step 2. Write the inequality as one quotient on the left and zero on the right.

The given function is$\frac{-2}{6x^2-13x+6}\le 0$

The inequality is in the correct form.

Factor the denominator.

Step 3. The equation of denominator is 6x2−13x+6

In the equation the product of first and last term is $36$and second term is $13$. So the possible factors are $4$and $9$.

$$\begin{gathered} =6x^2-(9+4)x+6 \\ =6x^2-9x-4x+6 \\ =3x(2x-3)-2(2x-3) \\ =(2x-3)(3x-2) \end{gathered}$$

$\frac{-2}{(2x-3)(3x-2)}\le 0$

Step 4. Now finding the critical points. 

The quotient is 0 when the numerator is 0.

Since the numerator is always $-2$, the quotient cannot be 0.

The quotient will be undefined when the denominator is zero.

$$\begin{gathered} 2x-3=03x-2=0 \\ x=\frac{3}{2}x=\frac{2}{3} \end{gathered}$$

Step 5. Use the critical points to divide the number line into intervals. 

The number line is divided into three intervals:

$(-\infty ,\frac{2}{3})(\frac{2}{3},\frac{3}{2})(\frac{3}{2},\infty )$

Step 6. Now choose the Interval   (-∞,23)

The number $0$is in the interval $(-\infty ,\frac{2}{3})$. Test $x=0$in the expression in the numerator and the denominator.

$\frac{-2}{(2x-3)(3x-2)}=\frac{-2}{(2\times 0-3)(3\times 0-2)}=\frac{-2}{(-3)(-2)}=\frac{-2}{6}=-\frac{1}{3}$

The quotient is marked negative in the interval $(-\infty ,\frac{2}{3})$.

Step 7. Now choose the Interval   (23,32)

The number 1 is in the interval $(\frac{2}{3},\frac{3}{2})$. Test $x=1$in the expression in the numerator and the denominator.

$\frac{-2}{(2x-3)(3x-2)}=\frac{-2}{(2\times 1-3)(3\times 1-2)}=\frac{-2}{(2-3)(3-2)}=\frac{-2}{(-1)·1}=2$

The quotient is marked positive in the interval $(\frac{2}{3},\frac{3}{2})$.

Step 8. Now choose the Interval   (32,∞)

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

$\frac{-2}{(2x-3)(3x-2)}=\frac{-2}{(2\times 2-3)(3\times 2-2)}=\frac{-2}{(4-3)(6-2)}=\frac{-2}{\left(1\right)\left(4\right)}=-\frac{1}{2}$

The quotient is marked negative in the interval $(\frac{3}{2},\infty )$.

Step 9. Determine the intervals where the inequality is correct. Write the solution in interval notation.  

We want the quotient to be less than or equal to zero, so the numbers in the intervals $(-\infty ,\frac{2}{3})$localid="1662723286410" $and\left(\frac{3}{2},\infty \right)$are solutions.

The solution in interval notation is localid="1662723300849" $(-\infty ,\frac{2}{3})\cup (\frac{3}{2},\infty )$.