Problem 50 - Beginning and Intermediate Algebra

Question

Solve each rational inequality. Graph the solution set and write the solution in interval notation. $$\frac{r}{r-7} \geq 0$$

Step-by-Step Solution

Answer

The solution to the rational inequality \(\frac{r}{r - 7} \geq 0\) is \((-\infty, 0) \cup (7, \infty)\).

Step 1: Identify critical points

To find the critical points of the rational inequality, find the values of r that make the numerator zero, and the values that make the denominator zero. For the numerator to be zero: $$r = 0$$ For the denominator to be zero: $$r - 7 = 0 \Rightarrow r = 7$$ The critical points are \(r = 0\) and \(r = 7\).

Step 2: Check intervals around critical points

To check the intervals around the critical points, we will choose a value from each interval and test the inequality. If it holds, the interval will be part of the solution. There are three intervals to test: 1. \(r < 0\) (for example, \(r = -1\)) 2. \(0 < r < 7\) (for example, \(r = 4\)) 3. \(r > 7\) (for example, \(r = 10\)) Test each of the representative values in the inequality \(\frac{r}{r - 7} \geq 0\): 1. For \(r = -1\), \(\frac{-1}{-1 - 7} = \frac{1}{6} \geq 0\) (Inequality holds) 2. For \(r = 4\), \(\frac{4}{4 - 7} = -\frac{4}{3} \geq 0\) (Inequality does not hold) 3. For \(r = 10\), \(\frac{10}{10 - 7} = \frac{10}{3} \geq 0\) (Inequality holds) The inequality holds in intervals \(r < 0\) and \(r > 7\).

Step 3: Graph the solution set

On a number line, shade the intervals where the inequality holds: 1. Shade from negative infinity to \(r = 0\), with an open circle at \(r = 0\) since the inequality is greater than or equal to zero. 2. Shade from \(r = 7\) to positive infinity, with an open circle at \(r = 7\) for the same reason.

Step 4: Write the solution in interval notation

To write the solution in interval notation, use the symbols for greater than or equal to (-∞) and less than or equal to (∞) with the appropriate intervals: Solution: \((-\infty, 0) \cup (7, \infty)\) The solution to the rational inequality \(\frac{r}{r - 7} \geq 0\) is \((-\infty, 0) \cup (7, \infty)\).

Key Concepts

Critical Points

Critical points are essential in understanding rational inequalities like \(\frac{r}{r-7} \geq 0\). To solve such inequalities, we first identify the critical points, which are the values that make the expression undefined or equal to zero. These points are found by analyzing both the numerator and the denominator of the fraction separately.

• Numerator Equals Zero: Set the numerator equal to zero. For \(\frac{r}{r-7}\), this means setting \(r = 0\).
• Denominator Equals Zero: Set the denominator equal to zero. For \(\frac{r}{r-7}\), this means \(r - 7 = 0\) or \(r = 7\).

After identifying these points, you can use them to divide the number line into intervals. It's within these intervals that we can test to see where the inequality holds. The critical points are essential as they help us determine the behavior of the inequality across different ranges of numbers.

Interval Notation

Interval notation is a way of expressing the set of solutions from an inequality. It concisely shows which intervals are part of the solution without listing all numbers individually.

• In writing interval notation, use parentheses \((\) or \()\) to denote that an endpoint is not included.
• Use brackets \([\) or \()]\) to denote that an endpoint is included.

For the inequality \(\frac{r}{r-7} \geq 0\), we determined the solution set does not include the endpoints (0 or 7), because the denominator cannot be zero in a rational expression. Thus, open intervals are used:

• The solution set is \((-\infty, 0) \cup (7, \infty)\).

This tells us that for this inequality, \(r\) can be any value less than 0 or greater than 7. Interval notation is a straightforward method to communicate the range of values, as seen here, where the union symbol \(\cup\) is used to combine multiple intervals.

Number Line Graphing

Number line graphing is a valuable way to visually represent the solution set for inequalities. It shows the intervals where the solution is valid.

• First, plot the critical points on the number line, here at \(r = 0\) and \(r = 7\).
• Use open circles to show that the endpoints are not included in the solution. This reflects that \(r\) cannot double as a denominator in our inequality \(\frac{r}{r-7}\).

Next, shade the sections of the number line that satisfy the inequality. In our case, these are the intervals \((-\infty, 0)\) and \((7, \infty)\). The shaded portion between the open circles distinctly shows the range of values that satisfy the inequality.

• Shading from negative infinity to 0, with an open circle at 0.
• Shading from 7 to positive infinity, with an open circle at 7.

Graphing on a number line provides a clear view of where values satisfy the inequality, complementing the written solution in interval notation.