Question

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

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

Short answer

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

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.

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

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

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

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

Step 3. Determine the critical points—the points where the rational expression will be zero or undefined. 

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 $4$.

Step 4. Use the critical points to divide the number line into intervals. 

The number line is divided into three intervals:

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

Step 5. Test a value in each interval. 

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 quotient positive in the interval $(-\infty ,-1)$.

Step 6. Now choose the Interval   (-1,4)

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 $(-1,4)$.

Step 7. Now choose the Interval   (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 $(4,\infty )$.

Step 8. 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 interval$(-1,4)$is the solution.

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