91Ó°ÊÓ

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

$\frac{3}{x^2−5x+4}<0$

The given function is$\frac{3}{x^2−5x+4}<0$

The inequality is in the correct form.

Factor the denominator.

In the equation the product of first and last term is $4$and second term is $5$. So the possible factors are $1$and $4$.

$$\begin{gathered} =x^2−(4+1)x+4 \\ =x^2−4x-x+4 \\ =x(x-4)-1(x-4) \\ =(x-1)(x-4) \end{gathered}$$

$\frac{3}{(x-1)(x-4)}<0$

The quotient is 0 when the numerator is 0.

Since the numerator is always 3, the quotient cannot be 0.

The quotient will be undefined when the denominator is zero.

$$\begin{gathered} x-1=0x-4=0 \\ x=1x=4 \end{gathered}$$

The number line is divided into three intervals:
$(-∞,1)(1,4)(4,∞)$

The number $0$is in the interval $(-∞,1)$. Test $x=0$in the expression in the numerator and the denominator.

$\frac{3}{(x-1)(x-4)}=\frac{3}{(0-1)(0-4)}=\frac{3}{(-1)(-4)}=\frac{3}{4}$

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

The number $2$is in the interval $(1,4)$. Test $x=2$in the expression in the numerator and the denominator.

$\frac{3}{(x-1)(x-4)}=\frac{3}{(2-1)(2-4)}=\frac{3}{1·(-2)}=\frac{3}{-2}$

The quotient is marked negative in the interval $(1,4)$.

The number $5$is in the interval $(4,∞)$. Test $x=5$in the expression in the numerator and the denominator.

$\frac{3}{(x-1)(x-4)}=\frac{3}{(5-1)(5-4)}=\frac{3}{4·1}=\frac{3}{4}$

The quotient is marked positive in the interval $(4,∞)$.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals $(-∞,1)$and $(4,∞)$are solutions.

The solution in interval notation is $(-∞,1)∪(4,∞)$.