Problem 66 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Triangle Inequality and the fact that \(0<|a|<|b| \Rightarrow 1 /|b|<1 /|a|\) to establish the following chain of inequalities. $$ \left|\frac{1}{x^{2}+3}-\frac{1}{|x|+2}\right| \leq \frac{1}{x^{2}+3}+\frac{1}{|x|+2} \leq \frac{1}{3}+\frac{1}{2} $$

Step-by-Step Solution

Answer

\(\left|\frac{1}{x^{2}+3}-\frac{1}{|x|+2}\right| \leq \frac{5}{6}\).

Step 1: Apply the Triangle Inequality

The triangle inequality states that for any real numbers \(x\) and \(y\), the inequality \(|x - y| \leq |x| + |y|\) holds. In this exercise, set \(x = \frac{1}{x^2 + 3}\) and \(y = \frac{1}{|x| + 2}\). Thus the inequality becomes \[ \left| \frac{1}{x^2+3} - \frac{1}{|x|+2} \right| \leq \frac{1}{x^2+3} + \frac{1}{|x|+2}. \] This is the statement provided by the triangle inequality rule without modifications. This establishes the first part of the chain of inequalities.

Step 2: Bound Each Individual Term

To bound each term \(\frac{1}{x^2+3}\) and \(\frac{1}{|x|+2}\), we need to consider their maximum values. Notice as \(x^2 + 3 \geq 3\) for all \(x\) and \(|x| + 2 \geq 2\) for all \(x\). Hence, \(\frac{1}{x^2+3} \leq \frac{1}{3}\) and \(\frac{1}{|x|+2} \leq \frac{1}{2}\).

Step 3: Summing the Boundaries

Now that we have bounds for each term, we add them together: \[ \frac{1}{x^2+3} + \frac{1}{|x|+2} \leq \frac{1}{3} + \frac{1}{2}. \] Calculate this addition: \[ \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}. \] This shows that the sum of the two expressions is less than or equal to \(\frac{5}{6}\).

Step 4: Conclusion: Establishing the Chain of Inequalities

Combine all parts from the previous steps: \[ \left| \frac{1}{x^{2}+3}-\frac{1}{|x|+2} \right| \leq \frac{1}{x^{2}+3}+\frac{1}{|x|+2} \leq \frac{5}{6}. \] This chain of inequalities is now fully established and verified.

Key Concepts

Triangle Inequality

The Triangle Inequality is a fundamental concept in mathematics and provides a way to compare the sum and difference of two numbers or expressions. It states that for any real numbers \(x\) and \(y\), the inequality \(|x - y| \leq |x| + |y|\) holds true.

This is particularly useful in scenarios where we want to measure the "distance" between different values, ensuring that this distance does not exceed the sum of their magnitudes.

In the exercise, by considering \(x = \frac{1}{x^2 + 3}\) and \(y = \frac{1}{|x| + 2}\), we applied this principle to establish the relationship: \ \[ \left| \frac{1}{x^2+3} - \frac{1}{|x|+2} \right| \leq \frac{1}{x^2+3} + \frac{1}{|x|+2}. \]

This tells us that the absolute difference between these two expressions is always less than or equal to their sum, which is a cornerstone in solving many mathematical inequalities.

Bounding Functions

Bounding functions is all about determining the upper or lower limits that a function can reach. In mathematical problems, understanding these bounds helps in narrowing down the possible values that a function can take.

For this exercise, we focus on bounding the expressions \(\frac{1}{x^2+3}\) and \(\frac{1}{|x|+2}\). Since \(x^2 + 3 \geq 3\) and \(|x| + 2 \geq 2\) for all \(x\), the inverse expressions follow:

• \(\frac{1}{x^2+3} \leq \frac{1}{3}\)
• \(\frac{1}{|x|+2} \leq \frac{1}{2}\)

These bounds were derived by finding the minimum values these denominators can achieve. The smaller the denominator, the larger the fraction, giving us clear boundaries for these expressions.

Mathematical Proofs

Mathematical proofs are a sequence of logical steps that demonstrate the truth of a statement. They are essential for validating results in mathematics.

In dealing with inequalities, proofs often use foundational principles such as the triangle inequality and bounding to derive the desired outcomes.

For the given problem, we provided a structured proof starting with the triangle inequality to form a preliminary bound: \[ \left| \frac{1}{x^{2}+3}-\frac{1}{|x|+2} \right| \leq \frac{1}{x^{2}+3}+\frac{1}{|x|+2} \]Then we used bounding techniques to refine this into another layer:\[ \frac{1}{x^{2}+3} + \frac{1}{|x|+2} \leq \frac{1}{3} + \frac{1}{2} \]Each step in this proof helps in solidifying the understanding of why the final inequality holds.

Chain of Inequalities

A chain of inequalities is a sequence of inequalities where the terms are linked in a relation, forming a continuous series of inequalities.

The given exercise demonstrates a chain of inequalities by first establishing an initial condition using the triangle inequality and then refining it using bounds. It begins with:

• \( \left| \frac{1}{x^2+3} - \frac{1}{|x|+2} \right| \leq \frac{1}{x^2+3} + \frac{1}{|x|+2} \)
• Followed by: \( \frac{1}{x^2+3} + \frac{1}{|x|+2} \leq \frac{1}{3} + \frac{1}{2} \)

These form the complete chain:\[ \left| \frac{1}{x^{2}+3}-\frac{1}{|x|+2} \right| \leq \frac{1}{x^{2}+3}+\frac{1}{|x|+2} \leq \frac{5}{6} \]Such chains are incredibly valuable in calculus and analysis, as they help break down complex relationships into more understandable segments, making it easier to see the big picture.