Problem 90 - Algebra and Trigonometry

Question

Solve each inequality in Exercises \(86-91\) using a graphing utility. $$ \frac{x+2}{x-3} \leq 2 $$

Step-by-Step Solution

Answer

The solutions to the inequality are \(x \in (-\infty , 3)\) and \(x \in (4, \infty)\)

Step 1: Arrange the inequality

First, the inequality needs to be in a form that can be easily graphed. So arrange the original inequality \(\frac{x+2}{x-3} \leq 2\) to the zero form by subtracting 2 from both sides: \(\frac{x+2}{x-3} - 2 \leq 0\). This can be further simplified to \(\frac{x+2 -2x +6}{x-3} \leq 0\), or \(\frac{4-x}{x-3} \leq 0\).

Step 2: Graph the inequality

Now, graph the function \(f(x)= \frac{4-x}{x-3}\) using a graphing tool. Pay close attention to the regions where \(f(x)\) is either above or below the x-axis. This will help identify the intervals that satisfy the inequality. Remember, within the graph, the function is less than or equal to zero where it is on or below the x-axis, and is greater than zero where it is above the x-axis.

Step 3: Identify the root

Next, find the roots of the inequality, these are the points where the function crosses the x-axis. The roots can also be found algebraically by setting \(f(x) = 0\) and solving for \(x\), in this case \(x = 4\). It's also important to notice where the function is undefined, this is when the denominator is zero. In this case, it is \(x = 3\). So we have two points to consider \((3, 4)\).

Step 4: Analyze the intervals

Using the x-values found in step 3, divide the number line into intervals. Test a number in each interval by substituting it into the inequality to see if it results in a true statement. Thereby, the graph shows that the function is below zero on two intervals: \((-\infty , 3)\) and \((4, \infty)\). Thus these are the solutions to the inequality.