Problem 60 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve the inequalities. $$ |2 x-1| \geq|x+1| $$

Step-by-Step Solution

Answer

The solutions are \((-\infty, -1) \cup [-1, 0] \cup [2, \infty)\).

Step 1: Understand the Inequality

The inequality given is \(|2x-1| \geq |x+1|\). This inequality involves absolute values, so we need to consider different cases based on where these expressions change sign.

Step 2: Identify Critical Points

The expressions inside the absolute values change signs at zero points. For \(2x-1\), it changes sign at \(x = \frac{1}{2}\). For \(x+1\), it changes sign at \(x = -1\). These critical points divide the number line into different intervals: \(x < -1\), \(-1 \leq x < \frac{1}{2}\), and \(x \geq \frac{1}{2}\).

Step 3: Solve for Each Interval

1. **Interval 1: \(x < -1\)** - Here, both expressions inside the absolute values are negative. Thus: \(-2x + 1 \geq -x - 1\) Solving gives: \(-2x + x \geq -1 - 1\) \(-x \geq -2\) \(x \leq 2\). This condition holds for all \(x < -1\).2. **Interval 2: \(-1 \leq x < \frac{1}{2}\)** - Here, \(2x-1\) is negative, \(x + 1\) is positive. \(-2x + 1 \geq x + 1\) Solving gives: \(-3x \geq 0\) \(x \leq 0\). This condition holds within the interval.3. **Interval 3: \(x \geq \frac{1}{2}\)** - Here, both expressions are non-negative. \(2x - 1 \geq x + 1\) \(2x - x \geq 1 + 1\) \(x \geq 2\). This condition holds for \(x \geq 2\).

Step 4: Combine Results

From the solved intervals:1. In interval 1 (\(x < -1\)), the solution is \((-\infty, -1)\).2. In interval 2 (\(-1 \leq x < \frac{1}{2}\)), the solution is \([-1, 0]\).3. In interval 3 (\(x \geq \frac{1}{2}\)), the solution is \([2, \infty)\).Combine all valid solutions: \((-\infty, -1) \cup [-1, 0] \cup [2, \infty)\).

Key Concepts

Absolute Values

Absolute values are a way of expressing the magnitude of a number without considering its sign. It is denoted by two vertical bars, for example, \(|x|\). The absolute value of a number is its distance from zero on the number line, regardless of direction. Thus, both \(|3|\) and \(|-3|\) are equal to 3.

When dealing with inequalities involving absolute values, it's essential to recognize the key property: \(|a| \geq b\) implies two cases, either \(a \geq b\) or \(-a \geq b\). This is because both positive and negative values could potentially satisfy the equation due to the absolute value, which can be likened to a mirror effect around zero.

In the original problem \(|2x-1| \geq |x+1|\), this property introduces multiple conditions that need to be tested across different intervals. These conditions arise due to points where expressions inside the absolute values change sign.

Critical Points

Critical points refer to values where the expressions inside the absolute values change from positive to negative or vice versa. Identifying these points is crucial because they determine the boundaries where the behavior of the inequality changes.

In the inequality \(|2x-1| \geq |x+1|\), we first find the critical points by setting the expressions inside the absolute values equal to zero:

• For \(2x-1 = 0\), solving gives us \(x = \frac{1}{2}\).
• For \(x+1 = 0\), solving gives us \(x = -1\).

These critical points, \(x = \frac{1}{2}\) and \(x = -1\), divide the number line into distinct sections. Each of these sections represents a different case that must be checked step by step to solve the inequality correctly. By understanding critical points, we can efficiently test the behavior of absolute value expressions within each resulting interval.

Number Line Intervals

Once you identify the critical points, you can create intervals on the number line. These intervals help us analyze the inequality by letting us determine how the inequality behaves within different ranges.

For the inequality \(|2x-1| \geq |x+1|\), the critical points \(x = -1\) and \(x = \frac{1}{2}\) create three intervals:

• **Interval 1:** \(x < -1\)
  In this interval, both expressions inside the absolute value are negative. Solving under this assumption helps us verify whether \(x\) satisfies the initial inequality.
• **Interval 2:** \(-1 \leq x < \frac{1}{2}\)
  Now, \(2x - 1\) turns negative, while \(x+1\) stays positive. These sign switches must be accounted for when deducing which solutions work.
• **Interval 3:** \(x \geq \frac{1}{2}\)
  Here, both expressions become non-negative because \(x\) is at or beyond the change point for each absolute value.

After evaluating each interval independently, we can then gather the valid results to construct the solution set for the original inequality. This method ensures accuracy when working with complex absolute value inequalities.