Problem 70 - Precalculus Student Solutions Manual 5th

Question

Solve each rational inequality. Write each solution set in interval notation. $$\frac{x+1}{x-4}>0$$

Step-by-Step Solution

Answer

The solution set is \((-\infty, -1) \cup (4, \infty)\).

Step 1: Find the critical points

The critical points are found by setting the numerator and denominator equal to 0. For the numerator: \(x + 1 = 0\) \(x = -1\) For the denominator: \(x - 4 = 0\) \(x = 4\)

Step 2: Determine the intervals

The critical points divide the number line into intervals. These intervals are \((-\infty, -1)\), \((-1, 4)\), and \((4, \infty)\).

Step 3: Test the sign of the expression in each interval

Test a point in each interval to determine the sign of \(\frac{x+1}{x-4}\).\((-\infty, -1)\): Test \(x = -2\). The expression \(\frac{-2+1}{-2-4}=\frac{-1}{-6}=\frac{1}{6} > 0\).\((-1, 4)\): Test \(x = 0\). The expression \(\frac{0+1}{0-4}=\frac{1}{-4}=-\frac{1}{4} < 0\).\((4, \infty)\): Test \(x = 5\). The expression \(\frac{5+1}{5-4}=\frac{6}{1}=6 > 0\).

Step 4: Determine the solution set

The expression \(\frac{x+1}{x-4} > 0\) in intervals where the test point resulted positive. Therefore, the solution set is \((-\infty, -1)\cup (4, \infty)\). The endpoints \(x = -1\) and \(x = 4\) are not included, because the inequality does not include equality.

Key Concepts

critical points

When solving rational inequalities, identifying the critical points is essential. Critical points are where the expression changes its sign and are found by setting both the numerator and the denominator equal to zero. For instance, consider the inequality \(\frac{x+1}{x-4}>0\). We first solve for the numerator and denominator: set \((x+1)=0\) which gives \(x=-1\), and set \(x-4=0\), which gives \(x=4\). These critical points, \(-1\) and \(4\), split the entire number line into distinct intervals, which we will examine next.

interval notation

After determining the critical points, the next step is to define the intervals these points create. In our example, the points \(-1\) and \(4\) divide the number line into three intervals: \((-\text{inf}, -1)\), \((-1, 4)\), and \((4, \text{inf})\). Interval notation is a precise way to describe continuous sets of values. Use brackets \[ \] to include an endpoint in the interval and parentheses \( \) to exclude it. Since in this example the inequality does not include equality (no \( \leq \) or \( \geq \)), we use parentheses to indicate that the critical points themselves are not part of the solution set.

testing intervals

Once intervals are defined, we need to test points from each interval to determine the sign of the rational expression inside these intervals. We choose any convenient number inside each interval and substitute it into the inequality. If we return to our example: for \((-\text{inf}, -1)\), we can pick \(x = -2\) and find \(\frac{-2+1}{-2-4} = \frac{-1}{-6} = \frac{1}{6} > 0\). For \((-1, 4)\), choosing \(x = 0\), we calculate \(\frac{0+1}{0-4} = \frac{1}{-4} = -\frac{1}{4} < 0\). Finally, for \((4, \text{inf})\), choosing \(x = 5\), we have \(\frac{5+1}{5-4} = \frac{6}{1} = 6 > 0\). Note how the sign of the expression tells us whether the inequality is satisfied in each interval.

numerator and denominator

When working with rational inequalities, always pay close attention to the numerator and the denominator separately. The critical points stem from setting both of these parts equal to zero: in \(\frac{x+1}{x-4} > 0\), \(x=-1\) from the numerator and \(x=4\) from the denominator change the behavior of the inequality. Remember, the denominator can't be zero because division by zero is undefined. Always exclude these points. Analyzing the conditions where the rational expression changes its sign around these critical points allows us to understand and solve the inequality accurately. So, breaking down the inequality into simpler parts (numerator and denominator) helps us carefully predict where the inequality holds true.