Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

For \(G(y)=1 /(y-1)\), find each value. (a) \(G(0)\) (b) \(G(0.999)\) (c) \(G(1.01)\) (d) \(G\left(y^{2}\right)\) (e) \(G(-x)\) (f) \(G\left(\frac{1}{x^{2}}\right)\)

Step-by-Step Solution

Answer

The values are: (a) -1, (b) -1000, (c) 100, (d) \(\frac{1}{(y-1)(y+1)}\), (e) \(\frac{1}{-x-1}\), (f) \(\frac{x^2}{1-x^2}\).

Step 1: Evaluate G(0)

To find \( G(0) \), substitute \( y = 0 \) into the function \( G(y) = \frac{1}{y-1}\). This gives: \( G(0) = \frac{1}{0-1} = \frac{1}{-1} = -1 \).

Step 2: Evaluate G(0.999)

Substitute \( y = 0.999 \) into \( G(y) = \frac{1}{y-1} \). This gives: \( G(0.999) = \frac{1}{0.999 - 1} = \frac{1}{-0.001} = -1000 \).

Step 3: Evaluate G(1.01)

Substitute \( y = 1.01 \) into \( G(y) = \frac{1}{y-1} \). This gives: \( G(1.01) = \frac{1}{1.01 - 1} = \frac{1}{0.01} = 100 \).

Step 4: Evaluate G(y^2)

Substitute \( y = y^2 \) into the function \( G(y) = \frac{1}{y-1} \). This gives \( G(y^2) = \frac{1}{y^2 - 1} \), which simplifies to \( rac{1}{(y-1)(y+1)} \).

Step 5: Evaluate G(-x)

Substitute \( y = -x \) into \( G(y) = \frac{1}{y-1} \). This gives \( G(-x) = \frac{1}{-x - 1} \).

Step 6: Evaluate G(\frac{1}{x^2})

Substitute \( y = \frac{1}{x^2} \) into \( G(y) = \frac{1}{y-1} \). This gives \( G\left(\frac{1}{x^2}\right) = \frac{1}{\frac{1}{x^2} - 1} \), which simplifies to \( \frac{1}{\frac{1-x^2}{x^2}} = \frac{x^2}{1-x^2} \).

Key Concepts

Function Evaluation

Function evaluation is like plugging numbers into a machine to see what comes out. When working with functions, the idea is to substitute the given input value into the function and calculate the result. Let's break it down using the function given in the exercise.

• The function is expressed as \( G(y) = \frac{1}{y-1} \).
• To evaluate \( G(0) \), plug in \( y=0 \) into the function. The math looks like this:\[G(0) = \frac{1}{0-1} = \frac{1}{-1} = -1\]
• Similarly, for \( G(0.999) \), substitute \( y = 0.999 \):\[G(0.999) = \frac{1}{0.999 - 1} = \frac{1}{-0.001} = -1000\]
• For \( G(1.01) \), substitute \( y = 1.01 \):\[G(1.01) = \frac{1}{1.01 - 1} = \frac{1}{0.01} = 100\]

This substitution process is what's referred to as function evaluation. It's crucial to carefully follow the order of operations during the calculations to get the correct result.

Rational Functions

A rational function is a type of function that represents the ratio of two polynomials. In simpler terms, it is like dividing one algebraic expression by another. Rational functions take the form:\[R(x) = \frac{P(x)}{Q(x)}\]where both \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \).

• In our exercise, \( G(y) = \frac{1}{y-1} \) is a rational function where the numerator is \( 1 \) (a constant polynomial) and the denominator is \( y-1 \).
• Rational functions are undefined wherever the denominator equals zero. For \( G(y) \), the function is undefined at \( y = 1 \) because dividing by zero is not possible.

Understanding rational functions and their properties, such as domains and asymptotes, helps in analyzing their behavior and graphing them effectively.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to make them more comprehensible. This skill is essential for evaluating functions, simplifying expressions, or solving equations. Let's look at some examples from the exercise.

• For \( G(y^2) \), substituting \( y = y^2 \) gives:\[G(y^2) = \frac{1}{y^2 - 1}\]This can be further factored as \( \frac{1}{(y-1)(y+1)} \), simplifying the expression and helping identify points where the function is undefined (\( y = 1 \) and \( y = -1 \)).
• In \( G\left(\frac{1}{x^2}\right) \), substitute \( y = \frac{1}{x^2} \):\[G\left(\frac{1}{x^2}\right) = \frac{1}{\frac{1}{x^2} - 1} = \frac{x^2}{1-x^2}\]This manipulation simplifies the original expression to help understand its behavior.

Mastering algebraic manipulation is crucial for unraveling complex expressions and making problem-solving manageable.