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Chapter 3: Problem 29

Solve each polynomial inequality in Exercises \(1-42\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ (x-1)(x-2)(x-3) \geq 0 $$

Short Answer

Expert verified
The solution to the inequality in interval notation is \([1,2] \cup [3,\infty)\).

Step by step solution

01

Factorize the polynomial

The given polynomial is \( (x-1)(x-2)(x-3) \geq 0 \). This polynomial is already factorized. The factors are \( x-1 \), \( x-2 \), and \( x-3 \).
02

Find the critical points

Critical points are the roots of the inequality and occur where the factor equals 0. So for each of the factors, solve \( x-1=0 \), \( x-2=0 \), and \( x-3=0 \) to get the critical points. Setting them equal to 0 and solving each factor yields \( x=1 \), \( x=2 \) and \( x=3 \).
03

Create a number line using the critical points

Place these critical points on a number line. The points divide the line into four intervals: \(-\infty, 1\), \(1, 2\), \(2, 3\), and \(3, \infty\).
04

Determine the sign of each interval

Choose a test point from each interval and substitute into the inequality to test the sign. We can choose 0, 1.5, 2.5, and 4. When \(x = 0\), the inequality \( (0-1)(0-2)(0-3) \geq 0 \) becomes \(-1*-2*-3 \geq 0\) which is false, hence \(-\infty, 1\) is not in the solution set. When \(x = 1.5\), \((1.5-1)(1.5-2)(1.5-3) \geq 0\) is true, so \(1, 2\) is in the solution set. When \(x = 2.5\), \((2.5-1)(2.5-2)(2.5-3) \geq 0\) is false, so \(2,3\) is not in the solution set. When \(x = 4\), \((4-1)(4-2)(4-3) \geq 0\) is true, so \(3, \infty\) is in the solution set.
05

Write final answer in interval notation

The final solution in interval notation which includes the end points (since the inequality includes \( \geq \)) and the solutions for each interval would then be \([1,2] \cup [3,\infty)\)

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