Chapter 0: Problem 22
Express the solution set of the given inequality in interval notation and sketch its graph. $$ (2 x+3)(3 x-1)(x-2)<0 $$
Short Answer
Expert verified
The solution set is \((-\frac{3}{2}, \frac{1}{3}) \cup (\frac{1}{3}, 2)\).
Step by step solution
01
Find the Zeros
To find the zeros of the inequality, set each factor equal 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 values partition the number line into intervals.
02
Determine the Intervals
Identify the intervals defined by the zeros: - \((-\infty, -\frac{3}{2})\) - \((-\frac{3}{2}, \frac{1}{3})\) - \((\frac{1}{3}, 2)\) - \((2, \infty)\).
03
Test Sign in Each Interval
Select test points from each interval and substitute them into the inequality \((2x+3)(3x-1)(x-2) < 0\):- For \((-\infty, -\frac{3}{2})\), choose \(x = -2\): \((2(-2)+3)(3(-2)-1)((-2)-2) > 0\).- For \((-\frac{3}{2}, \frac{1}{3})\), choose \(x = 0\): \((2(0)+3)(3(0)-1)(0-2) < 0\).- For \((\frac{1}{3}, 2)\), choose \(x = 1\): \((2(1)+3)(3(1)-1)(1-2) < 0\).- For \((2, \infty)\), choose \(x = 3\): \((2(3)+3)(3(3)-1)(3-2) > 0\).
04
Identify the Solution Set
Based on testing:- The inequality is negative for \(x \in (-\frac{3}{2}, \frac{1}{3}) \cup (\frac{1}{3}, 2)\).- The zeros \(-\frac{3}{2}\), \(\frac{1}{3}\), and \(2\) are not included since the inequality is strict \((< 0)\).
05
Sketch the Graph
Draw a number line and mark the zeros: \(-\frac{3}{2}, \frac{1}{3}, \) and \(2\). Shade the intervals \((-\frac{3}{2}, \frac{1}{3})\) and \((\frac{1}{3}, 2)\) to indicate where the inequality is true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Zeros of a function
Finding zeros of a function is like discovering the places where a graph touches or crosses the x-axis. To find these zeros, you set each factor of the function equal to zero and solve for the variable. In the context of inequalities, zeros play a crucial role in determining the points where the sign of the function might change. For instance, with the function
- \( (2x+3)(3x-1)(x-2) \)
- \( 2x + 3 = 0 \): Solving gives \( x = -\frac{3}{2} \).
- \( 3x - 1 = 0 \): Solving gives \( x = \frac{1}{3} \).
- \( x - 2 = 0 \): Solving gives \( x = 2 \).
Interval notation
Interval notation provides a concise way to describe sets of numbers, particularly solutions to inequalities. It uses parentheses \(( )\) and brackets \([ ]\) to indicate if endpoints are included or excluded. When you solve an inequality, such as \((2x+3)(3x-1)(x-2)<0\),the solution set finds expression in interval notation. Here's how you do it:
- Parentheses \(( )\) indicate that an endpoint is not included in the solution.
- Brackets \([ ]\) mean the endpoint is included.
Graphing inequalities
Graphing inequalities involves plotting solutions on a number line or coordinate plane. It visually represents the set of values making an inequality true. In this example, the inequality \((2x+3)(3x-1)(x-2)<0\) translates to certain intervals on a number line. First, plot the zeros found earlier at \( x = -\frac{3}{2}, \frac{1}{3}, \) and \(2\).Here’s how you graphically show the solution:
- Draw a number line.
- Mark zeros as distinct points. Since the inequality is strict (<), use open circles to illustrate zeros aren't included.
- Shade the intervals where the inequality holds true: \((-\frac{3}{2}, \frac{1}{3})\) and \((\frac{1}{3}, 2)\).