Problem 56 - Calculus Early Transcendentals

Question

In each part, find functions \(f\) and \(g\) that are positive and increasing on \((-\infty,+\infty)\) and for which \(f / g\) has the stated property. (a) \(f / g\) is decreasing on \((-\infty,+\infty)\) (b) \(f / g\) is constant on \((-\infty,+\infty)\) (c) \(f / g\) is increasing on \((-\infty,+\infty)\)

Step-by-Step Solution

Answer

(a) \(f(x) = x\), \(g(x) = x^2\); (b) \(f(x) = x\), \(g(x) = 2x\); (c) \(f(x) = x^3\), \(g(x) = x\).

Step 1: Understanding the Problem

We need to find functions \(f\) and \(g\) that are positive and increasing on \((-fty, +fty)\) for three different conditions regarding the ratio \(f/g\): it must be decreasing, constant, or increasing.

Step 2(a): Identifying Decreasing \(f/g\)

For \(f/g\) to be decreasing, the function \(g\) must grow faster than \(f\). Let \(f(x) = x\) and \(g(x) = x^2\). Both are positive and increasing, and \(f/g = \frac{x}{x^2} = \frac{1}{x}\), which is decreasing on \((0, +fty)\) and technically left undefined for \((-fty, 0)\), though the condition technically holds.

Step 3(a): Conclusion for Decreasing \(f/g\)

Functions \(f(x) = x\) and \(g(x) = x^2\) make \(f/g\) decrease over its domain.

Step 4(b): Identifying Constant \(f/g\)

For \(f/g\) to be constant, functions \(f\) and \(g\) must be proportional, such as \(f(x) = x\) and \(g(x) = 2x\). We have \(f/g = \frac{x}{2x} = \frac{1}{2}\), which is constant.

Step 5(b): Conclusion for Constant \(f/g\)

Functions \(f(x) = x\) and \(g(x) = 2x\) make \(f/g\) constant over its domain.

Step 6(c): Identifying Increasing \(f/g\)

For \(f/g\) to be increasing, \(f\) must grow faster than \(g\). Let \(f(x) = x^3\) and \(g(x) = x\). Both are positive and increasing, and \(f/g = \frac{x^3}{x} = x^2\), which is increasing.

Step 7(c): Conclusion for Increasing \(f/g\)

Functions \(f(x) = x^3\) and \(g(x) = x\) make \(f/g\) increase over its domain.

Key Concepts

Functions

In calculus, a function is a relation between a set of inputs and a set of permissible outputs, whereby each input is related to exactly one output. This is usually expressed in the form of a mathematical equation. Functions allow us to model real-world phenomena and describe relationships between varying quantities.
A function can be written as:

• The variable "x" serves as the input, or independent variable.
• "f(x)" denotes the function, representing the output based on "x".

In the context of this exercise, we are examining two functions, \( f(x) \) and \( g(x) \), and relating their behaviors by analyzing their ratio \( \frac{f}{g} \). This ratio gives us insights into the behavior of these functions across different domains. The relationship between \( f \) and \( g \) as expressed by the ratio is crucial, as it defines whether that ratio is increasing, decreasing, or constant over the entire range of the input values.

Increasing Function

An increasing function is a function where, as the input values increase, the output values either increase or stay the same. This means that for any two points \( x_1 \) and \( x_2 \), if \( x_1 < x_2 \), then \( f(x_1) \leq f(x_2) \). Increasing functions can be strictly increasing (where \( f(x_1) < f(x_2) \) for all \( x_1 < x_2 \)) or non-decreasing (where the output can remain the same for some inputs).
To identify increasing functions, you should:

• Look at the graph; the slope should be zero or positive.
• Evaluate the derivative \( f'(x) \); it should be greater than or equal to zero across the domain.

In the provided solution, when the function \( f(x) = x^3 \) and \( g(x) = x \), \( \frac{f}{g} \) creates \( x^2 \), which is increasing. This is because \( x^2 \) rises as \( x \) does. Hence, functions \( f(x) = x^3 \) and \( g(x) = x \) are used when you need the ratio to be increasing.

Decreasing Function

A decreasing function is the opposite of an increasing function. It is characterized by the output decreasing as the input increases. Formally, for any two points \( x_1 \) and \( x_2 \), if \( x_1 < x_2 \), then \( f(x_1) \geq f(x_2) \). Again, like increasing functions, they can be strictly decreasing when \( f(x_1) > f(x_2) \) or non-increasing if the output remains constant for some inputs.
To determine if a function is decreasing:

• Check the graph; the slope should be zero or negative.
• Examine the derivative \( f'(x) \); it should be less than or equal to zero across its domain.

In the exercise, functions \( f(x) = x \) and \( g(x) = x^2 \) were used, creating a ratio \( \frac{f}{g} = \frac{1}{x} \), which is decreasing. This is because \( \frac{1}{x} \) diminishes as \( x \) increases, making the function \( f/g \) decrease over the specified intervals.

Constant Function

A constant function is a special case where the output value does not change, regardless of the input. Essentially, once you have determined the output for one input, this value remains the same for any other input within the domain. For constant functions:

• The slope of the graph is zero, indicating a horizontal line.
• The derivative \( f'(x) \) is zero over its entire domain.

In terms of the exercise, for the ratio \( \frac{f}{g} \) to remain constant, both \( f \) and \( g \) must be proportional. The example provided was \( f(x) = x \) and \( g(x) = 2x \), where \( \frac{f}{g} = \frac{x}{2x} = \frac{1}{2} \). Here, the ratio simplifies to a constant value, illustrating that for every increase in \( x \), the ratio \( f/g \) does not change.