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

$\frac{4x-1}{x-4}\le 1$

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

$\frac{4x-1}{x-4}\le 1$

The given function is$\frac{4x-1}{x-4}\le 1$

Subtract by $1$on both side

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

Simplifying the function

$$\begin{gathered} \frac{4x-1}{x-4}-\frac{1(x-4)}{x-4}\le 0 \\ \frac{4x-1-1(x-4)}{x-4}\le 0 \\ \frac{4x-1-x+4}{x-4}\le 0 \\ \frac{3x+3}{x-4}\le 0 \end{gathered}$$

The rational expression will be zero when the numerator is zero. Since $3x+3=0$when $x=-1$, then $-1$is a critical point.

The rational expression will be undefined when the denominator is zero. Since $x-4=0$when $x=4$, then $4$is a critical point.

The critical points are $-1$and $4$.

The number line is divided into three intervals:

$(-\infty ,-1)(-1,4)(4,\infty )$

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.

$\mathit{I}\mathit{n}\mathit{t}\mathit{e}\mathit{r}\mathit{v}\mathit{a}\mathit{l}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}$

The number $-2$is in the interval $(-\infty ,-1)$. Test $-2$in the expression in the numerator and the denominator.

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

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

$\frac{3x+3}{x-4}=\frac{3\times 1+3}{1-4}=\frac{3+3}{-3}=\frac{6}{-3}=-2$

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

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

$\frac{3x+3}{x-4}=\frac{3\times 5+3}{5-4}=\frac{15+3}{1}=18$

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

We want the quotient to be less than or equal to zero, so the numbers in the interval$(-1,4)$is the solution.

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