Problem 40 - Beginning and Intermediate Algebra

Question

Solve each inequality. Graph the solution set and write the solution in interval notation. $$(t+2)(4 t-7)(5 t-1) \geq 0$$

Step-by-Step Solution

Answer

The solution set for the inequality \((t+2)(4t-7)(5t-1) \geq 0\) is \((-∞, -2] \cup [\frac{7}{4}, ∞)\).

Step 1: Find when the inequality equals 0

To find the points where the inequality is equal to 0, we will set each of the factors to 0: (1) \(t + 2 = 0\) (2) \(4t - 7 = 0\) (3) \(5t - 1 = 0\) Now we'll solve for t in each equation: (1) \(t = -2\) (2) \(t = \frac{7}{4}\) (3) \(t = \frac{1}{5}\) These are our critical points: \(-2, \frac{7}{4},\) and \(\frac{1}{5}\).

Step 2: Determine the intervals

Now, we will divide the real number line into intervals based on our critical points. We will have four intervals: Interval 1: \(-\infty\) to \(-2\) Interval 2: \(-2\) to \(\frac{1}{5}\) Interval 3: \(\frac{1}{5}\) to \(\frac{7}{4}\) Interval 4: \(\frac{7}{4}\) to \(\infty\)

Step 3: Test a number within each interval

Now, we'll test a number within each interval to see if it makes the inequality true: Interval 1: \(-3\) \((-3 + 2)(4(-3) - 7)(5(-3) - 1) < 0\), which is true. Interval 2: \(0\) \((0 + 2)(4(0) - 7)(5(0) - 1) > 0\), which is false. Interval 3: \(1\) \((1 + 2)(4(1) - 7)(5(1) - 1) > 0\), which is false. Interval 4: \(2\) \((2 + 2)(4(2) - 7)(5(2) - 1) < 0\), which is true.

Step 4: Graph the solution set

Now, let's graph the solution set, which includes the intervals 1 and 4, including the critical points: [-----(-∞, -2] U [7/4, ∞)-----]

Step 5: Write the solution in interval notation

Finally, we write the solution set in interval notation: \((-∞, -2] \cup [\frac{7}{4}, ∞)\)

Key Concepts

Critical Points

Critical points in an inequality are values that make the equation equal to zero. Identifying these points is crucial as they help divide the number line into different intervals, guiding us to determine the solution set.

For the inequality \((t+2)(4t-7)(5t-1) \geq 0\), we first find the critical points by setting each factor to zero:

• \(t + 2 = 0\) leads to \(t = -2\)
• \(4t - 7 = 0\) leads to \(t = \frac{7}{4}\)
• \(5t - 1 = 0\) leads to \(t = \frac{1}{5}\)

These calculated critical points, \(-2, \frac{7}{4}, \frac{1}{5}\), serve as boundaries for evaluating the inequality over different intervals on the number line.

Interval Notation

Once we identify the critical points, we divide the number line into intervals that we need to test. This process helps determine where the inequality holds true.

The intervals for our inequality are:

• Interval 1: \((-\infty, -2)\)
• Interval 2: \((-2, \frac{1}{5})\)
• Interval 3: \((\frac{1}{5}, \frac{7}{4})\)
• Interval 4: \((\frac{7}{4}, \infty)\)

Testing a value from each interval helps us determine if the inequality is satisfied in those intervals, which can then be recorded using interval notation. For example, in this scenario, the inequality is satisfied in Interval 1 and Interval 4. Hence, we record the solution as \((-\infty, -2] \cup [\frac{7}{4}, \infty)\) in interval notation.

Inequality Graphing

Graphing inequalities involves plotting the intervals where the inequality holds true, which provides a visual representation of the solution set.

For graphing \((t+2)(4t-7)(5t-1) \geq 0\), intervals found from testing include \((-\infty, -2]\) and \([\frac{7}{4}, \infty)\).

• Closed Interval: A filled dot on the graph indicates that the endpoint belongs to the solution set, as in \((-\infty, -2]\).
• Open Interval: An open dot implies the endpoint is not part of the solution, generally seen with notations where equality is not involved.

This graphical approach allows us to easily see which portions of the number line satisfy the inequality conditions, making it a useful tool for visual learners.