Problem 92 - Precalculus Student Solutions Manual 5th

Question

Solve each rational inequality. Write each solution set in interval notation.4 $$\frac{(9 x-11)(2 x+7)}{(3 x-8)^{3}}>0$$

Step-by-Step Solution

Answer

The solution set is \[(-\frac{7}{2}, \frac{11}{9}) \cup (\frac{8}{3}, \infty)\]

Step 1: Identify the critical points

Set the numerator and denominator equal to zero to find the critical points:\[(9x - 11)(2x + 7) = 0 \implies x = \frac{11}{9}, x = -\frac{7}{2}\]\[(3x - 8)^3 = 0 \implies x = \frac{8}{3} \]

Step 2: Determine test intervals

The critical points divide the number line into four intervals: \((-\infty, -\frac{7}{2}), (-\frac{7}{2}, \frac{11}{9}), (\frac{11}{9}, \frac{8}{3}), (\frac{8}{3}, \infty)\)

Step 3: Test each interval

Choose a test point in each interval to determine the sign of the expression:1. For \((-\infty, -\frac{7}{2})\), choose \(x = -4\): \[ \frac{(9(-4)-11)(2(-4)+7)}{(3(-4)-8)^3} > 0 \implies \frac{(-36-11)(-8+7)}{(-12-8)^3} = \frac{(-47)(-1)}{(-20)^3} > 0 \implies False \]2. For \((-\frac{7}{2}, \frac{11}{9})\), choose \(x = 0\): \[ \frac{(9(0)-11)(2(0)+7)}{(3(0)-8)^3} > 0 \implies \frac{(-11)(7)}{(-8)^3} = \frac{(-77)}{-512} > 0 \implies True \]3. For \((\frac{11}{9}, \frac{8}{3})\), choose \(x = 2\): \[ \frac{(9(2)-11)(2(2)+7)}{(3(2)-8)^3} > 0 \implies \frac{(18-11)(4+7)}{(6-8)^3} = \frac{(7)(11)}{-2^3} = \frac{77}{-8} > 0 \implies False \]4. For \((\frac{8}{3}, \infty)\), choose \(x = 3\): \[ \frac{(9(3)-11)(2(3)+7)}{(3(3)-8)^3} > 0 \implies \frac{(27-11)(6+7)}{(9-8)^3} = \frac{(16)(13)}{1} > 0 \implies True \]

Step 4: Write the solution set

The solution set consists of the intervals where the expression is positive: \[(-\frac{7}{2}, \frac{11}{9}) \cup (\frac{8}{3}, \infty)\]

Key Concepts

critical points

In rational inequalities, critical points are the values where the numerator or the denominator of the rational expression equals zero.
To find these points, set each factor of the numerator and denominator equal to zero, and solve for the variable.
In our specific example, we identify the critical points by solving the equations for the factors:

• \((9x - 11) = 0\), yielding \(x = \frac{11}{9}\).
• \((2x + 7) = 0\), yielding \(x = -\frac{7}{2}\).
• \((3x - 8)^3 = 0\), yielding \(x = \frac{8}{3}\).

These critical points are essential because they break the number line into different intervals, which will be examined individually to solve the inequality. Also, critical points where the denominator is zero are not included in the solution as they make the rational expression undefined.

test intervals

After finding the critical points, the next step is to determine the intervals that we will test.
By placing the critical points on a number line, we can divide the number line into different segments. For the example with critical points, we get the intervals:

• \((-fty, -\frac{7}{2})\)
• \((-\frac{7}{2}, \frac{11}{9})\)
• \((\frac{11}{9}, \frac{8}{3})\)
• \((\frac{8}{3}, fty)\)

These intervals will be tested to determine if the rational expression is greater than zero in each one. To do this, we select a test point in each interval. A test point is any number within the interval. For example, choosing \(x = -4\) in \((-fty, -\frac{7}{2})\) and evaluating the rational inequality at this point.

interval notation

Once we know which intervals satisfy the inequality, we express the solution using interval notation.
Interval notation is a way to describe a set of numbers along the number line. It includes:

• Parentheses \(()\) for intervals that do not include the endpoint (open intervals).
• Brackets \([]\) for intervals that include the endpoint (closed intervals).

In our example, the rational expression is positive in the intervals \((-\frac{7}{2}, \frac{11}{9})\) and \((\frac{8}{3}, fty)\). Therefore, we write the solution set as:

• \[(-\frac{7}{2}, \frac{11}{9}) \cup (\frac{8}{3}, fty)\]

Here, \(\cup\) denotes the union of the two intervals, indicating all the values within both intervals combined.