Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the solution sets of the given inequalities. $$ \left|\frac{1}{x}-3\right|>6 $$

Step-by-Step Solution

Answer

The solution set is \((-\frac{1}{3}, 0) \cup (0, \frac{1}{9})\).

Step 1: Understand the Inequality

The given inequality is \( \left|\frac{1}{x}-3\right|>6 \). This means we need to find the values of \( x \) such that the expression inside the absolute value is greater than 6 or less than -6.

Step 2: Remove the Absolute Value

The inequality \( \left|\frac{1}{x}-3\right|>6 \) can be split into two separate inequalities: \( \frac{1}{x} - 3 > 6 \) and \( \frac{1}{x} - 3 < -6 \).

Step 3: Solve the First Inequality

For \( \frac{1}{x} - 3 > 6 \), add 3 to both sides to get \( \frac{1}{x} > 9 \). Taking the reciprocal and reversing the inequality (as the reciprocal of a positive number switches the inequality), we get \( x < \frac{1}{9} \).

Step 4: Solve the Second Inequality

For \( \frac{1}{x} - 3 < -6 \), add 3 to both sides to get \( \frac{1}{x} < -3 \). Taking the reciprocal and reversing the inequality, we find \( x > -\frac{1}{3} \).

Step 5: Combine the Solutions

The solutions to the two inequalities are \( x < \frac{1}{9} \) and \( x > -\frac{1}{3} \). Considering the domain of the original function \( \frac{1}{x} \) (which excludes \( x = 0 \)), the final solution is \( -\frac{1}{3} < x < 0 \) and \( 0 < x < \frac{1}{9} \).

Step 6: Write the Final Solution

The solution to the inequality \( \left|\frac{1}{x}-3\right|>6 \) is \( x \in (-\frac{1}{3}, 0) \cup (0, \frac{1}{9}) \).

Key Concepts

Absolute Value Inequalities

Absolute value inequalities involve expressions within absolute value bars. These bars \(|...|\) represent the distance a number is from zero on a number line, without considering direction. In our exercise, the inequality is \(|\frac{1}{x} - 3| > 6\), which means we're interested in all the values where the distance between \(\frac{1}{x} - 3\) and zero is greater than 6. This splits into two situations: \(\frac{1}{x} - 3 > 6\) or \(\frac{1}{x} - 3 < -6\) because the absolute value could be greater than 6 in either direction. These represent two potential ranges for \(x\).

Solution Sets

A solution set consists of all values that satisfy a given mathematical condition, such as an inequality. For the inequality \(|\frac{1}{x} - 3| > 6\), identifying a solution set involves solving both parts of the inequality separately: \(\frac{1}{x} - 3 > 6\) and \(\frac{1}{x} - 3 < -6\).

• First solution: \(x < \frac{1}{9}\)
• Second solution: \(x > -\frac{1}{3}\)

We combine these solutions, considering that \(x\) cannot be zero since \(\frac{1}{x}\) is undefined at \(x = 0\). The final solution set is represented using interval notation, accounting for the domain limitations.

Reciprocal and Inequality Reversal

When working with inequalities involving a reciprocal, it's important to remember that taking the reciprocal of both sides of an inequality reverses the inequality's direction, provided both sides are positive. In our problem:

• For \(\frac{1}{x} > 9\), the reciprocal gives \(x < \frac{1}{9}\).
• For \(\frac{1}{x} < -3\), the reciprocal results in \(x > -\frac{1}{3}\).

Always check the signs and remember that this step changes the inequality's sense.

Inequality Domain

Understanding the domain in which an inequality is defined is critical when solving it. In the context of the inequality \(|\frac{1}{x} - 3| > 6\), the function \(\frac{1}{x}\) dictates that \(x\) cannot equal zero, as it would make the expression undefined. This restriction is crucial as it divides the number line into intervals where the inequality's solutions can be found. Recognizing these restrictions helps in assembling the final solution set accurately and prevents incorrect solutions from being considered.

Interval Notation

Interval notation is a way of expressing solution sets of inequalities, using intervals to describe where the solutions lie on a number line. In this exercise, the solution set \(-\frac{1}{3} < x < 0\) and \(0 < x < \frac{1}{9}\) is expressed in interval notation as \(-\frac{1}{3}, 0 \) and \(0, \frac{1}{9}\). The use of parentheses indicates that the endpoints are not included in the solution set (open intervals), reflecting that \(x\) cannot be \(-\frac{1}{3}\), 0, or \(\frac{1}{9}\). This notation provides a compact, clear way to communicate which values of \(x\) satisfy the inequality.