91Ó°ÊÓ

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

$\frac{4x−3}{x−3}â‰\yen 2$

The given function is$\frac{4x−3}{x−3}â‰\yen 2$

Subtract by $2$on both side

$$\begin{gathered} \frac{4x−3}{x−3}-2â‰\yen 2-2 \\ \frac{4x−3}{x−3}-2â‰\yen 0 \end{gathered}$$

Simplifying the function

$$\begin{gathered} \frac{4x−3}{x−3}-\frac{2(x−3)}{x−3}â‰\yen 0 \\ \frac{4x−3-2(x−3)}{x−3}â‰\yen 0 \\ \frac{4x−3-2x+6}{x−3}â‰\yen 0 \\ \frac{2x+3}{x−3}â‰\yen 0 \end{gathered}$$

The rational expression will be zero when the numerator is zero. Since $2x+3=0$when $x=-\frac{3}{2}$, then $-\frac{3}{2}$is a critical point.

The rational expression will be undefined when the denominator is zero. Since $x-3=0$when $x=3$, then $3$is a critical point.

The critical points are $-\frac{3}{2}$and $3$.

The number line is divided into three intervals:

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

To find the sign of each factor in an interval, we choose any point in that interval and use it as a test point. Any point in the interval will give the expression the same sign, so we can choose any point in the interval.

$\mathit{I}\mathit{n}\mathit{t}\mathit{e}\mathit{r}\mathit{v}\mathit{a}\mathit{l}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{∞}\mathbf{,}\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{)}$

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

$\frac{2x+3}{x−3}=\frac{2×(-2)+3}{-2−3}=\frac{-4+3}{-5}=\frac{-1}{-5}=\frac{1}{5}$

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

$\frac{2x+3}{x−3}=\frac{2×1+3}{1−3}=\frac{2+3}{-2}=-\frac{5}{2}$

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

The number $4$is in the interval $(3,∞)$. Test $x=4$in the expression in the numerator and the denominator.

$\frac{2x+3}{x−3}=\frac{2×4+3}{4−3}=\frac{8+3}{1}=11$

The quotient is marked negative in the interval $(3,∞)$.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals $(-∞,-\frac{3}{2})$and $(3,∞)$are solutions.

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