Problem 43

Question

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

Step-by-Step Solution

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Answer

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

1Step 1: Understand the Inequality

We have an inequality involving an absolute value: $ \left| \frac{1}{x} - 3 \right| > 6 $. This inequality means that the expression $ \frac{1}{x} - 3 $ can be either greater than 6 or less than -6.

2Step 2: Split the Absolute Value Inequality

Since an absolute value inequality $ \left| A \right| > B $ splits into two separate inequalities, we have:1. $ \frac{1}{x} - 3 > 6 $ and2. $ \frac{1}{x} - 3 < -6 $.

3Step 3: Solve the First Inequality

For $ \frac{1}{x} - 3 > 6 $, add 3 to both sides:$ \frac{1}{x} > 9 $.To solve $ \frac{1}{x} > 9 $, multiply both sides by $ x \cdot x $, but remember $ x eq 0 $:$ 1 > 9x $,or equivalently, $ x < \frac{1}{9} $ since multiplying by a negative reverses the inequality.

4Step 4: Solve the Second Inequality

For $ \frac{1}{x} - 3 < -6 $, add 3 to both sides:$ \frac{1}{x} < -3 $.To solve $ \frac{1}{x} < -3 $, multiply both sides by $ x \cdot x $, ensuring $ x eq 0 $:$ 1 < -3x $,or equivalently, $ x > -\frac{1}{3} $.

5Step 5: Combine the Solutions and Consider Domain

The solutions to the inequalities $ x < \frac{1}{9} $ and $ x > -\frac{1}{3} $ must exclude 0 because $ x eq 0 $ (the constraint for the fraction $ \frac{1}{x} $).Thus, the combined solution considering the domain is:$(-\frac{1}{3}, 0) \cup (0, \frac{1}{9})$.

Key Concepts

absolute value inequalityrational expressionsinequality intervalssolving inequalities

absolute value inequality

In mathematics, when we see an absolute value inequality, it involves the absolute distance of a number from zero on the number line. The absolute value inequality like $ \left| A \right| > B $ indicates that the expression $ A $ can be more than $ B $ in both the positive and negative directions on the number line.

In the context of our problem, we have $ \left| \frac{1}{x} - 3 \right| > 6 $. This inequality splits into two separate inequalities:

$ \frac{1}{x} - 3 > 6 $
$ \frac{1}{x} - 3 < -6 $

The rationale is based on how absolute values behave, focusing on both the greater than positive and less than the negative. Understanding this splitting will help us visualize that the original inequality is exploring two possible regions on the number line.

rational expressions

A rational expression involves a quotient where both the numerator and the denominator are polynomials. In our exercise, the expression $ \frac{1}{x} $ is a rational expression where 1 is the numerator and $ x $ is the denominator.

The main challenge with rational expressions is their undefined values when the denominator equals zero. Thus, for $ \frac{1}{x} $, $ x eq 0 $ is an important restriction to remember. It means we exclude $ x = 0 $ from our solution since division by zero is undefined.

When solving inequalities involving rational expressions, we also need to multiply by the square of $ x $ to clear the fraction. This requires careful attention to sign changes as the inequality might need to be reversed if $ x $ is negative.

inequality intervals

Inequality intervals represent parts of the number line that satisfy a given inequality. When we subtract or add numbers while solving the inequalities, we're essentially finding these intervals.

For $ \frac{1}{x} - 3 > 6 $, solving leads to $ x < \frac{1}{9} $. Conversely, the inequality $ \frac{1}{x} - 3 < -6 $ solves as $ x > -\frac{1}{3} $. These represent two separate intervals on the number line.

Since we cannot use $ x = 0 $, our valid numbers are all within the intervals:

From $-\frac{1}{3}$ to just before 0
From just after 0 to $ \frac{1}{9} $

Understanding and plotting these intervals on the number line helps visualize where the solutions fall, showing that our combined solution is $ (-\frac{1}{3}, 0) \cup (0, \frac{1}{9}) $.

solving inequalities

Solving inequalities involves finding all values that satisfy the inequality conditions. It often includes using several operations: addition, subtraction, multiplication, or division of both sides of the inequality. But there's a crucial thing to remember—the inequality symbol can change direction.

For instance, when we solve $ \frac{1}{x} > 9 $, multiplying both sides by $ x^2 $ (since $ x eq 0 $) gives $ 1 > 9x $, leading to $ x < \frac{1}{9} $. The opposite inequality, $ \frac{1}{x} < -3 $, when similarly solved, gives $ x > -\frac{1}{3} $.

Consequently, solving these inequalities involves these steps:

Isolating the variable to one side
Carefully assessing sign changes in the inequality
Ensuring any solutions respect the domain (here, $ x eq 0 $)

This detailed stepwise approach guarantees that all possible solutions validly satisfy the inequality conditions.

Problem 43

Other exercises in this chapter

Problem 43

. Find the area of the sector of a circle of radius 5 centime ters and central angle 2 radians (see Problem 42).

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Problem 43

Let $f(x+y)=f(x)+f(y)$ for all $x$ and $y$. Prove that there is a number $m$ such that $f(t)=m t$ for all rational numbers t. Hint: First decide what

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Problem 43

Since $0.199999 \ldots=0.200000 \ldots$ and $0.399999 \ldots=$ $0.400000 \ldots$ (see Problems 41 and 42), we see that certain rational numbers have two d

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Problem 44

Find a formula for $f^{-1}(x)$ and then verify that $f^{-1}(f(x))=x$ and $f\left(f^{-1}(x)\right)=x$ $$ f(x)=\left(\frac{x^{3}+2}{x^{3}+1}\right)^{5} $$

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