Problem 23 - Calculus with Concepts in Calculus

Question

Solve the inequality. $$ (x-1)(x+1 / 2) \geq 0 $$

Step-by-Step Solution

Answer

\([-\frac{1}{2}, 1] \cup (1, \infty)\)

Step 1: Identify Critical Points

To solve the inequality \((x-1)(x+\frac{1}{2}) \geq 0\), first find the critical points by setting each factor to zero. For \(x-1 = 0\), \(x = 1\). For \(x+\frac{1}{2} = 0\), \(x = -\frac{1}{2}\). Hence, the critical points are \(x = 1\) and \(x = -\frac{1}{2}\).

Step 2: Divide the Number Line

Use the critical points to divide the number line into intervals: \((-\infty, -\frac{1}{2})\), \((-\frac{1}{2}, 1)\), and \((1, \infty)\).

Step 3: Test Intervals

Choose test points from each interval to determine the sign of the expression \((x-1)(x+\frac{1}{2})\). Test point \(x = -1\) from \((-\infty, -\frac{1}{2})\): the expression is \((x-1)(x+\frac{1}{2}) = (-1-1)(-1+\frac{1}{2}) = (-2)(-0.5) = 1\). Test point \(x = 0\) from \((-\frac{1}{2}, 1)\): the expression is \((x-1)(x+\frac{1}{2}) = (0-1)(0+\frac{1}{2}) = (-1)(0.5) = -0.5\). Test point \(x = 2\) from \((1, \infty)\): the expression is \((x-1)(x+\frac{1}{2}) = (2-1)(2+0.5) = (1)(2.5) = 2.5\).

Step 4: Include Critical Points and Determine Solution

Check the critical points in the original inequality: \At \(x = -\frac{1}{2}\), \((x-1)(x+\frac{1}{2}) = (-1.5)(0) = 0\), which satisfies the inequality.\At \(x = 1\), \((x-1)(x+\frac{1}{2}) = (0)(1.5) = 0\), which also satisfies the inequality.\Thus, the solution to the inequality \((x-1)(x+\frac{1}{2}) \geq 0\) is \(x \in [-\frac{1}{2}, 1] \cup (1, \infty)\).

Key Concepts

Critical Points

Critical points are the values of \( x \) where the expression equals zero. In the inequality \((x-1)(x+\frac{1}{2}) \geq 0\), we identify these points by setting each factor of the expression to zero:

• For \(x-1=0\), we solve for \(x\) to get \(x = 1\).
• For \(x+\frac{1}{2}=0\), solving for \(x\) gives \(x = -\frac{1}{2}\).

These critical points break the number line into separate intervals. By finding where the expression equals zero, we can focus on these specific points to conduct further analysis and determine which intervals satisfy the inequality.

Number Line Test

The number line test involves dividing the number line into intervals based on the critical points. For our inequality, \(x = 1\) and \(x = -\frac{1}{2}\) split the number line into three segments:

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

This visual division helps us better understand the nature of the sign of the expression in each segment. It's essential to know which segments satisfy the inequality. Using this structured approach ensures no interval is overlooked during the process.

Sign Analysis

Sign analysis requires evaluating the expression within each interval divided by the critical points. By choosing a test point from each interval, you can determine if the product of the factors is positive or negative:

• For \((-\infty, -\frac{1}{2})\), let's take \(x = -1\). The value is positive \[(-2)(-0.5) = 1\].
• For \((-\frac{1}{2}, 1)\), select \(x = 0\). The value is negative \[(-1)(0.5) = -0.5\].
• In \((1, \infty)\), choose \(x = 2\). The value is positive \[(1)(2.5) = 2.5\].

This analysis quickly tells us which sections satisfy the inequality: the intervals that yield a positive result or zero meet the condition \((x-1)(x+\frac{1}{2}) \geq 0\).

Interval Notation

Interval notation is a compact way of describing the solution set that satisfies the inequality. After performing the sign analysis, we found that solutions exist in certain intervals, which can be expressed concisely:

• The critical point \(x = -\frac{1}{2}\) when tested, yields zero, thus it is included: \([-\frac{1}{2}, 1]\).
• The interval from \(1\) to infinity, \((1, \infty)\), also results in positive values.This whole solution is written as \([-\frac{1}{2}, 1] \cup (1, \infty)\).

The use of square brackets \([]\) denotes inclusivity of endpoints. Parentheses \(\) represent non-inclusivity, signaling that a particular endpoint is not part of the solution. This notation efficiently represents the complete range of \(x\) values fulfilling the original inequality.