Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Express the solution set of the given inequality in interval notation and sketch its graph. $$ (2 x+3)(3 x-1)(x-2)<0 $$

Step-by-Step Solution

Answer

The solution set is \((-\infty, -\frac{3}{2}) \cup (\frac{1}{3}, 2)\).

Step 1: Find the Zero Points

To solve the inequality \[(2x + 3)(3x - 1)(x - 2) < 0\]we begin by setting each factor to zero:1. \(2x + 3 = 0\) gives \(x = -\frac{3}{2}\)2. \(3x - 1 = 0\) gives \(x = \frac{1}{3}\)3. \(x - 2 = 0\) gives \(x = 2\)These are the points where the expression changes sign.

Step 2: Determine the Intervals

The zero points divide the number line into four intervals:1. \((-\infty, -\frac{3}{2})\)2. \((-\frac{3}{2}, \frac{1}{3})\)3. \((\frac{1}{3}, 2)\)4. \((2, \infty)\)

Step 3: Test Each Interval

Select a test point from each interval and substitute it into the expression:- For \((-\infty, -\frac{3}{2})\), choose \(x = -2\): \((2(-2)+3)(3(-2)-1)((-2)-2) = -\cdot -\cdot - = -\)- For \((-\frac{3}{2}, \frac{1}{3})\), choose \(x = 0\): \((2(0)+3)(3(0)-1)(0-2) = +\cdot -\cdot - = +\)- For \((\frac{1}{3}, 2)\), choose \(x = 1\): \((2(1)+3)(3(1)-1)(1-2) = +\cdot +\cdot - = -\)- For \((2, \infty)\), choose \(x = 3\): \((2(3)+3)(3(3)-1)(3-2) = +\cdot +\cdot + = +\)The expression is negative in the intervals \((-\infty, -\frac{3}{2})\) and \((\frac{1}{3}, 2)\).

Step 4: Express the Solution in Interval Notation

Since we want where the expression is less than zero, the solution in interval notation is:\[(-\infty, -\frac{3}{2}) \cup (\frac{1}{3}, 2)\]

Step 5: Sketch the Graph

To sketch the graph of \[(2x + 3)(3x - 1)(x - 2)\], mark the zero points \(x = -\frac{3}{2}\), \(x = \frac{1}{3}\), and \(x = 2\) on the x-axis. Indicate the intervals where the expression is negative by shading the intervals \((-\infty, -\frac{3}{2})\) and \((\frac{1}{3}, 2)\) on the x-axis. The function crosses the x-axis at these zero points and changes sign at each of them.

Key Concepts

Zero Points

When solving an inequality like \((2x + 3)(3x - 1)(x - 2) < 0\), it's crucial to first find the zero points. The zero points are the x-values where each factor of the equation becomes zero. Finding these points is the first step, as they indicate where the expression might change sign.

To find these zero points:

• Set each factor equal to zero and solve for x. This results in the equations:
  • \(2x + 3 = 0\) gives \(x = -\frac{3}{2}\)
  • \(3x - 1 = 0\) gives \(x = \frac{1}{3}\)
  • \(x - 2 = 0\) gives \(x = 2\)

The zero points are \(x = -\frac{3}{2}\), \(x = \frac{1}{3}\), and \(x = 2\). These are the places on the x-axis where our inequality changes direction.

Interval Notation

Interval notation is a way of representing a set of numbers between a lower and upper bound. In the context of solving inequalities, interval notation provides a concise way to describe where an expression is true or false. Once the zero points are determined, they divide the number line into different sections or intervals. These intervals help assess where the inequality is satisfied.

With \((2x + 3)(3x - 1)(x - 2) <0\), these zero points divide the x-axis into four intervals:

• \((-\infty, -\frac{3}{2})\)
• \((-\frac{3}{2}, \frac{1}{3})\)
• \((\frac{1}{3}, 2)\)
• \((2, \infty)\)

By testing each interval, we identify which intervals make the expression negative. This defines our solution set in interval notation: \((-\infty, -\frac{3}{2}) \cup (\frac{1}{3}, 2)\). It includes numbers from both segments that satisfy the inequality.

Graph Sketching

Graph sketching involves visually representing the behavior of an algebraic expression on a coordinate axis. Once the zero points and solution intervals are known, the next step is to sketch the graph. It's an effective way to illustrate where the inequality holds.

Here's how you sketch \((2x + 3)(3x - 1)(x - 2)\):

• Plot zero points on the x-axis: \(x = -\frac{3}{2}\), \(x = \frac{1}{3}\), and \(x = 2\).
• Indicate the intervals where the expression is negative by shading the sections \((-\infty, -\frac{3}{2})\) and \((\frac{1}{3}, 2)\).
• The graph crosses the x-axis at these points, flipping sign at each zero.

By sketching, students can see how graph changes across different intervals, helping solidify the concept of inequality solving.

Sign Changes

Sign changes are critical to solving inequalities because they determine where an expression shifts from positive to negative, or vice versa. Understanding these changes helps pinpoint the intervals where the original inequality holds true.

As you evaluate the intervals:

• Choose test points within each interval. For instance, \(x = -2\) in \((-\infty, -\frac{3}{2})\), \(x = 0\) in \((-\frac{3}{2}, \frac{1}{3})\), \(x = 1\) in \((\frac{1}{3}, 2)\), and \(x = 3\) in \((2, \infty)\).
• Substitute each into the inequality and compute the expression's sign.
  • Negative values indicate where the inequality is true.

The expression is negative across the intervals \((-\infty, -\frac{3}{2})\) and \((\frac{1}{3}, 2)\), making them part of the solution. This demonstrates where the coating layers of algebra show thickness of less than zero.