Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the solution sets of the given inequalities. $$ \left|2+\frac{5}{x}\right|>1 $$

Step-by-Step Solution

Answer

The solution is \((-5, 0) \cup (0, \frac{5}{3})\).

Step 1: Set Up the Inequality

First, we handle the absolute value inequality \( \left|2 + \frac{5}{x}\right| > 1 \). This means we have to solve both of its possible cases separately: \( 2 + \frac{5}{x} > 1 \) and \( 2 + \frac{5}{x} < -1 \).

Step 2: Solve Case 1

For the first case, solve \( 2 + \frac{5}{x} > 1 \):1. Subtract 2 from both sides: \( \frac{5}{x} > -1 \).2. Solve for \( x \) by multiplying both sides by \( x \) (note: \( x eq 0 \)): \( 5 > -x \).3. Divide by -1, which reverses the inequality: \( x > -5 \).

Step 3: Solve Case 2

For the second case, solve \( 2 + \frac{5}{x} < -1 \):1. Subtract 2 from both sides: \( \frac{5}{x} < -3 \).2. Solve for \( x \) by multiplying both sides by \( x \) (note: \( x eq 0 \)): \( 5 < -3x \).3. Divide by -3, which reverses the inequality: \( x < \frac{5}{3} \).

Step 4: Determine Critical Points and Exclusions

Identify any critical points or values for which the original inequality is undefined. Since the inequality involves a fraction with a variable in the denominator, \( x eq 0 \) is a critical point.

Step 5: Analyze Solution Sets

Combine the solutions found:- From **Case 1**: \( x > -5 \)- From **Case 2**: \( x < \frac{5}{3} \)Since combining these solves both sides of the original inequality, the actual solutions are the union of the sets, excluding \( x = 0 \) (where the original expression is undefined).

Step 6: Write the Final Solution Set

Both solution intervals intersect, and we must exclude 0, so the final solution in interval notation is \((-\infty, 0) \cup (0, \frac{5}{3}) \cap (-5, \infty)\), which simplifies to \((-5, 0) \cup (0, \frac{5}{3})\).

Key Concepts

Absolute Value

The concept of absolute value is crucial when dealing with inequalities involving expressions like \( \left|2 + \frac{5}{x}\right| > 1 \). The absolute value of a number is its distance from zero on the number line, regardless of direction. This means it is always non-negative. When an absolute value inequality is set up, it splits into two separate inequalities. For instance, \( \left|2 + \frac{5}{x}\right| > 1 \) splits into two cases:

• \( 2 + \frac{5}{x} > 1 \)
• \( 2 + \frac{5}{x} < -1 \)

In simple terms, you're looking at the scenarios where the expression inside the absolute value is either more than 1 or less than -1. This is because the absolute value reflects the magnitude, not the sign, of the expression.

Solution Sets

Solution sets are collections of values that satisfy a given inequality. They are derived after solving both cases in an absolute value inequality. For the exercise, solving the inequality \( \left|2 + \frac{5}{x} \right| > 1 \) provides two separate solution sets:

• From case 1, where \( 2 + \frac{5}{x} > 1 \), the solution is \( x > -5 \).
• From case 2, \( 2 + \frac{5}{x} < -1 \), the solution is \( x < \frac{5}{3} \).

By combining these solutions, we determine the values of \( x \) that satisfy either condition. However, certain critical points like \( x = 0 \) must be excluded as they make the expression undefined. Thus, the overall solution set is a combination of results from both cases, minus any undefined values.

Interval Notation

Interval notation is a simplified way to write the solution sets of inequalities. It captures the range of values between numbers easily. After solving the inequality \( \left|2 + \frac{5}{x}\right| > 1 \), we arrive at the conditions \( x > -5 \) and \( x < \frac{5}{3} \), excluding \( x = 0 \).In interval notation, the solution set is recorded as \((-5, 0) \cup (0, \frac{5}{3})\). This notation specifies:

• \((-5, 0)\): all numbers greater than -5 and less than 0.
• \((0, \frac{5}{3})\): all numbers greater than 0 and less than \(\frac{5}{3}\).

The use of union (\(\cup\)) indicates that values from both intervals are included in the solution. Parentheses denote strict inequalities, meaning endpoints are not included, which aligns with the problem requirements.

Critical Points

Critical points are values where an expression becomes undefined or behaves differently. In the context of the inequality \( \left|2 + \frac{5}{x}\right| > 1 \), a critical point to consider is \( x = 0 \).Since \( x \) appears in the denominator, it cannot be zero because division by zero is undefined. Identifying this critical point ensures we exclude it from the solution set. Critical points often indicate where a function's output suddenly changes, or becomes infinite.In our case, excluding \( x = 0 \) is crucial for forming the accurate solution set as \((-5, 0) \cup (0, \frac{5}{3})\). This approach ensures no undefined or non-feasible values are included.