Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

\text { 28. } y=\left(x^{6}+2\right) /\left(8 x^{2}\right), 1 \leq x \leq 3

Step-by-Step Solution

Answer

The range of \( y \) is \( \left[ \frac{3}{8}, 10.15 \right] \).

Step 1: Understand the Expression

First, identify the expression given: \( y = \frac{x^6 + 2}{8x^2} \). This is a rational expression where the numerator is \( x^6 + 2 \) and the denominator is \( 8x^2 \).

Step 2: Simplify the Expression

To simplify, divide both terms of the numerator by the denominator: \( y = \frac{x^6}{8x^2} + \frac{2}{8x^2} \). Simplifying further, \( \frac{x^6}{8x^2} = \frac{x^{6-2}}{8} = \frac{x^4}{8} \), and \( \frac{2}{8x^2} = \frac{1}{4x^2} \). So, the expression simplifies to \( y = \frac{x^4}{8} + \frac{1}{4x^2} \).

Step 3: Define the Domain

The problem states the domain as \( 1 \leq x \leq 3 \). This means we need to consider the behavior of \( y \) only for these values of \( x \).

Step 4: Analyze at Boundary Points

Calculate \( y \) at the boundary points of the domain. For \( x = 1 \), \( y = \frac{1^4}{8} + \frac{1}{4(1)^2} = \frac{1}{8} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8} \). For \( x = 3 \), \( y = \frac{3^4}{8} + \frac{1}{4(3)^2} = \frac{81}{8} + \frac{1}{36} \). Simplifying the second term, \( \frac{1}{36} = \frac{1}{36} \), so \( y = \frac{81}{8} + \frac{1}{36} \). Convert to a common denominator to add the fractions, resulting in \( \frac{729}{72} + \frac{2}{72} = \frac{731}{72} \).

Step 5: Evaluate Expression Behavior

Between \( x = 1 \) and \( x = 3 \), \( y = \frac{x^4}{8} + \frac{1}{4x^2} \) is a smoothly increasing function due to the dominance of the \( \frac{x^4}{8} \) term, which grows as \( x \) increases. Verify using the first derivative that this function is increasing.

Step 6: Conclusion - Range of \( y \)

Given that \( y \) is increasing on \( 1 \leq x \leq 3 \), the minimum value is \( y(1) = \frac{3}{8} \) and the maximum value is \( y(3) = \frac{731}{72} \) or about \( 10.15 \). Thus, the range of \( y \) is \( \left[ \frac{3}{8}, 10.15 \right] \).

Key Concepts

Rational Expression

A rational expression is something you'll often encounter in algebra and calculus. It's essentially a fraction where both the numerator and the denominator are polynomials. In other words, it's of the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \).
In the given exercise, the function \( y = \frac{x^6 + 2}{8x^2} \) represents a rational expression. Here:

• The numerator is \( P(x) = x^6 + 2 \), and
• The denominator is \( Q(x) = 8x^2 \).

To work with rational expressions, simplifying them is crucial. Simplification is done by breaking down both the numerator and the denominator to find common factors and reducing them.
For example, in \( \frac{x^6 + 2}{8x^2} \), we simplify to \( \frac{x^4}{8} + \frac{1}{4x^2} \) by dividing both terms in the numerator by the denominator.

Function Analysis

Analyzing a function involves understanding its behavior over its domain, such as where it increases or decreases, its maximum and minimum values, and how it changes.
In our problem, we're dealing with the function \( y = \frac{x^4}{8} + \frac{1}{4x^2} \). This analysis reveals the nature of the function between the boundaries provided by the domain.
For example:

• We note that \( \frac{x^4}{8} \) becomes larger as \( x \) increases, indicating that \( y \) is increasing within our limits of \( x \).
• The second term \( \frac{1}{4x^2} \) becomes smaller as \( x \) grows, because it is divided by a larger square.

To confirm that the function is increasing or decreasing, calculate the first derivative \( y' \) of the function and analyze its sign over the domain. If \( y' > 0 \), \( y \) is increasing.
This is valuable as it helps determine critical points and the overall "shape" of the graph of \( y \).

Domain and Range

The domain and range of a function are key concepts in understanding how a function behaves.
The domain of a function is all the input values \( x \) for which the function is defined. In this exercise, the domain is given explicitly: \( 1 \leq x \leq 3 \). This means we're only interested in analyzing the function on this interval.

• Knowing the domain helps us predict the values for which the expression won’t be undefined or lead to errors.
• It's critical to ensure the denominator never equals zero in rational expressions, although in this case, the domain avoids \( x = 0 \).

The range is all possible output values \( y \) the function can produce. From our problem, once we examine the values at the boundaries \( x = 1 \) and \( x = 3 \), we calculate:

• For \( x = 1 \), \( y = \frac{3}{8} \)
• For \( x = 3 \), \( y \approx 10.15 \)

Thus, the range of \( y \) is \( \left[ \frac{3}{8}, 10.15 \right] \). This tells us all possible outputs when \( x \) is within the given domain. It's like the 'reach' of the function.