Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve each inequality. Express your solution in interval notation. (a) \(1.99<\frac{1}{x}<2.01\) (b) \(2.99<\frac{1}{x+2}<3.01\)

Step-by-Step Solution

Answer

(a) \((0.4975, 0.5025)\); (b) \((-1.6678, -1.6656)\)

Step 1: Solve the First Inequality

We want to solve the inequality \(1.99<\frac{1}{x}<2.01\). This is a compound inequality that implies \(1.99<\frac{1}{x}\) and \(\frac{1}{x}<2.01\). To solve \(\frac{1}{x} > 1.99\), we take the reciprocal on both sides, remembering to switch the inequality sign: \(x < \frac{1}{1.99}\). To solve \(\frac{1}{x} < 2.01\), take the reciprocal on both sides, thus changing the inequality sign again: \(x > \frac{1}{2.01}\). Therefore, the solution is \(\frac{1}{2.01} < x < \frac{1}{1.99}\).

Step 2: Express the Solution in Interval Notation for Part (a)

Now that we have \(\frac{1}{2.01} < x < \frac{1}{1.99}\), compute the actual values for \(\frac{1}{2.01}\) and \(\frac{1}{1.99}\). Approximating these gives us: \(x \in (0.4975, 0.5025)\). Therefore, the solution is expressed in interval notation as \((0.4975, 0.5025)\).

Step 3: Solve the Second Inequality

We now solve the inequality \(2.99<\frac{1}{x+2}<3.01\). This splits into two inequalities: \(2.99<\frac{1}{x+2}\) and \(\frac{1}{x+2}<3.01\). For \(2.99 < \frac{1}{x+2}\), we again take reciprocals and switch inequality signs: \(x+2 < \frac{1}{2.99}\). For \(\frac{1}{x+2} < 3.01\), reciprocate and switch inequality signs: \(x+2 > \frac{1}{3.01}\). So, simplify to \(x < \frac{1}{2.99} - 2\) and \(x > \frac{1}{3.01} - 2\).

Step 4: Express the Solution in Interval Notation for Part (b)

Calculate the values: \(\frac{1}{2.99} \approx 0.3344\) and \(\frac{1}{3.01} \approx 0.3322\). Thus, \(x \in (0.3322 - 2, 0.3344 - 2) = (-1.6678, -1.6656)\). Therefore, for the second inequality, the solution in interval notation is \((-1.6678, -1.6656)\).

Key Concepts

Interval Notation

Interval notation is a way to describe the set of solutions to an inequality in a concise manner. It uses parentheses and brackets to show which numbers are included in the solution set. In mathematics, parentheses "(" and ")" are used to indicate that an endpoint is not included in the interval, known as an open interval. For instance,

• \((a, b)\) means all numbers greater than \(a\) and less than \(b\), but not \(a\) or \(b\) themselves.

On the other hand, brackets "[" and "]" mean that the endpoint is included, referred to as a closed interval.

• \([a, b]\) would include both \(a\) and \(b\).

For the given exercises, we encountered open intervals. For example, the solution to part (a) was expressed as

• \((0.4975, 0.5025)\), where neither endpoint is part of the solution set.

Compound Inequalities

Compound inequalities involve two separate inequalities that must both be true simultaneously. These can be joined by the word 'and', indicating that the solution must satisfy both parts of the inequality. In the given examples:

• The task was to solve inequalities such as \(1.99 < \frac{1}{x} < 2.01\).
• This can be broken down into two separate inequalities: \(1.99 < \frac{1}{x}\) and \(\frac{1}{x} < 2.01\).

Both these conditions must be met. To solve them, we often take steps like finding common solutions or intervals that satisfy both inequalities. This might involve solving each inequality individually and then finding where their solutions overlap. It's this common overlap that gives us the final solution range. For example, in part (a), solutions for \(x\) were found and expressed as \(\frac{1}{2.01} < x < \frac{1}{1.99}\), which were further calculated as an interval in the solution.

Reciprocal Inequalities

Reciprocal inequalities arise when the variable is in the denominator. For example,

• In the inequality \(\frac{1}{x}\), solving involves using reciprocals.

It's important to know that taking the reciprocal of each side of an inequality also requires switching the direction of the inequality sign. This is a critical step, because the original inequality's order (greater than/less than) must be reversed after calculating reciprocals. For example,

• \( \frac{1}{x} > 1.99\) becomes \(x < \frac{1}{1.99}\).

Without changing the direction of the sign, the inequality would represent a fundamentally different set of solutions. Understanding this flip is vital when working through these problems. In our exercise, such techniques were applied to solve both inequalities given in parts (a) and (b), making sure the solutions were correctly represented.