Problem 37 - Beginning and Intermediate Algebra

Question

Solve each inequality. Graph the solution set and write the solution in interval notation. $$(j-7)(j-5)(j+9) \geq 0$$

Step-by-Step Solution

Answer

The solution to the inequality \((j-7)(j-5)(j+9) \geq 0\) is given by the union of two intervals, which are \((-\infty, -9)\) and \((5, 7)\). In interval notation, the solution can be written as \((-\infty, -9) \cup (5, 7)\).

Step 1: Find the critical points

The critical points are the values of \(j\) at which the inequality is equal to zero. To find them, we equate each factor to zero: \(j-7=0\) => \(j=7\) \(j-5=0\) => \(j=5\) \(j+9=0\) => \(j=-9\) Now we know that the critical points are -9, 5, and 7.

Step 2: Determine the intervals satisfying the inequality

From the critical points, we can divide the \(j\)-axis into four intervals: \((-\infty, -9)\), \((-9, 5)\), \((5, 7)\), and \((7, \infty)\). We will now need to check each interval to see if it satisfies the inequality. For each interval, choose a test value and substitute it into the inequality: 1. Interval \((-\infty, -9)\): let's take the test value \(j = -10\): \((-10 - 7)(-10 - 5)(-10+9) < 0\), the inequality holds true for this interval. 2. Interval \((-9, 5)\): let's take the test value \(j = 0\): \((0 - 7)(0 - 5)(0 + 9) > 0\), the inequality does not hold true for this interval. 3. Interval \((5, 7)\): let's take the test value \(j = 6\): \((6 - 7)(6 - 5)(6 + 9) < 0\), the inequality holds true for this interval. 4. Interval \((7, \infty)\): let's take the test value \(j = 8\): \((8 - 7)(8 - 5)(8+9) > 0\), the inequality does not hold true for this interval. Based on the above analysis, we find that the intervals \((-\infty, -9)\) and \((5, 7)\) satisfy the inequality.

Step 3: Graph the solution set

To graph the solution set, we draw a number line representing the \(j\)-axis, and plot the critical points (-9, 5, and 7) on it. Then, we shade the intervals which satisfy the inequality, which were \((-\infty, -9)\) and \((5, 7)\), and parentheses are used for the endpoint values.

Step 4: Write the solution in interval notation

Finally, we write the solution to the inequality in interval notation, which is the union of the two solution intervals: \((-\infty, -9) \cup (5, 7)\)

Key Concepts

Critical Points

In algebra, particularly when dealing with polynomial inequalities, critical points are key to understanding where the inequality changes its behavior. The critical points are the values of the variable where the expression equals zero. For example, for the expression \((j-7)(j-5)(j+9) \geq 0\), we set each factor to zero to find the critical points. This gives us three critical points:

• \(j = 7\) from \(j - 7 = 0\)
• \(j = 5\) from \(j - 5 = 0\)
• \(j = -9\) from \(j + 9 = 0\)

These values divide the number line into different intervals. The critical points are where the inequality can potentially change its direction from being positive to negative or vice versa. In solving the inequality, understanding these points is essential to test each interval they create.

Interval Notation

Interval notation is a concise way to express a set of numbers along the number line. It is often used to denote the range of solutions for inequalities. In the exercise, after identifying the critical points \(-9, 5, \text{and} 7\), these divide the number line into distinct intervals:

• \((-\infty, -9)\)
• \((-9, 5)\)
• \((5, 7)\)
• \((7, \infty)\)

We then test a random number within each interval to determine if they satisfy the inequality. Intervals that satisfy the inequality are written in interval notation to show the range of values for which the inequality holds true. In this case, the solution in interval notation would be \((-\infty, -9) \cup (5, 7)\). Parentheses indicate that the endpoints are not included in the solution set because the inequality is non-strict (\(\geq\)).

Solution Set

The solution set in mathematics refers to the collection of all possible solutions that satisfy a given inequality, equation, or system. For inequalities, once we determine the intervals that satisfy the inequality, they form the solution set. In this case, testing each interval against the inequality \((j-7)(j-5)(j+9) \geq 0\), show that only the intervals \((-\infty, -9)\) and \((5, 7)\) meet the condition. Consequently, the solution set is expressed as a union of these intervals: \((-\infty, -9) \cup (5, 7)\). This tells us precisely where the values of \(j\) satisfy the inequality. To communicate this graphically, you plot these intervals on a number line, shading these specific sections. This method not only provides a visual representation but also assists in understanding the behavior of the inequality across the number line.