/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 Determine whether the statements... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the statements are true or false. If a statement is false, find functions for which the statement fails to hold. 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 \((-\infty,+\infty)\).

Short Answer

Expert verified
(a) True; (b) True; (c) True.

Step by step solution

01

Analyze Part (a) Statement

The statement is that \( \frac{f}{g} \) is decreasing on \((-fty, +fty)\). For this to hold, \( \frac{f(x)}{g(x)} \) must be less than \( \frac{f(y)}{g(y)} \) for all \( x < y \). This implies that the derivative \( \frac{d}{dx} \left( \frac{f}{g} \right) < 0 \) for all \( x \). Let's try \( f(x) = e^x \) and \( g(x) = e^{2x} \).
02

Confirm Part (a) with Derivatives

Calculate \( \frac{f}{g} = \frac{e^x}{e^{2x}} = e^{-x} \). The derivative is \( \frac{d}{dx}(e^{-x}) = -e^{-x} \) which is negative on \((-fty, +fty)\). Thus, \( \frac{f}{g} \) is decreasing. This makes the statement true for these functions.
03

Analyze Part (b) Statement

The statement is that \( \frac{f}{g} \) is constant. If \( \frac{f}{g} = c \), a constant, then \( f = cg \). Let's try \( f(x) = g(x) = e^x \).
04

Confirm Part (b) by Calculation

For \( f(x) = e^x \) and \( g(x) = e^x \), the quotient \( \frac{f}{g} = \frac{e^x}{e^x} = 1 \), which is constant on \((-fty, +fty)\). Therefore, the statement is true.
05

Analyze Part (c) Statement

The statement is that \( \frac{f}{g} \) is increasing. This means \( \frac{d}{dx} \left( \frac{f}{g} \right) > 0 \). Consider \( f(x) = e^{2x} \) and \( g(x) = e^x \).
06

Confirm Part (c) with Derivatives

Calculate \( \frac{f}{g} = \frac{e^{2x}}{e^x} = e^x \). The derivative is \( \frac{d}{dx}(e^x) = e^x \), which is positive on \((-fty, +fty)\). Thus, \( \frac{f}{g} \) is increasing and the statement is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Functions Analysis
In the realm of calculus, functions analysis is the study of mathematical functions and their varying behaviors within a defined interval. By analyzing functions, mathematicians and students can determine characteristics such as whether a function is increasing, decreasing, or staying constant over a specific range.
When assessing functions, one often examines the relationship between two functions expressed as a quotient, such as \( f/g \). Here, the functions \( f \) and \( g \) are considered over their entire domain, \((-fty, +fty)\).
The main goal is to understand how the composite function \( \frac{f}{g} \) behaves. To do this, it is crucial to examine the function properties like continuity and differentiability. By analyzing these properties, we can gain insight into the behavior of the function quotient in various scenarios: decreasing, constant, or increasing.
In the presented exercise, students are tasked with determining whether specific statements about the function ratio \( \frac{f}{g} \) are true, by analyzing the derivative of the quotient and its sign across the domain.
Derivatives
The derivative of a function indicates how the function's value changes as its input changes. It provides vital information about the rate of change of the function and is a cornerstone in calculus for understanding functions' behaviors. Calculating derivatives involves applying rules such as the power rule, product rule, quotient rule, and chain rule.
The solution for the exercise heavily relies on the concept of derivatives. When examining the function \( \frac{f}{g} \), its derivative helps us determine how the quotient changes across its domain.
For a function \( \frac{f}{g} \), we utilize the quotient rule to find its derivative. The quotient rule states that:
  • \( \frac{d}{dx} \left( \frac{f}{g} \right) = \frac{g \cdot f' - f \cdot g'}{g^2} \)
This derivative is analyzed to determine whether the quotient is increasing, decreasing, or constant:
  • A positive derivative means the quotient is increasing.
  • A negative derivative means it is decreasing.
  • A zero derivative across an interval means it is constant.
By checking the sign of \( \frac{d}{dx} \left( \frac{f}{g} \right) \), one can confidently assert the behavior of the function \( \frac{f}{g} \). This is applied in the solutions for each part of the exercise.
Decreasing and Increasing Functions
Understanding whether functions are decreasing or increasing is crucial in calculus, as it tells us about the direction of change of the function values. An increasing function is one where, for any two values \( x < y \), the function value at \( x \) is less than at \( y \). Conversely, a decreasing function has a function value that is higher at \( x \) than at \( y \).
In other words, increasing functions have a positive rate of change, and decreasing functions have a negative rate of change. This is directly tied to the derivative:
  • If \( \frac{d}{dx}(f) > 0 \), then \( f \) is increasing.
  • If \( \frac{d}{dx}(f) < 0 \), then \( f \) is decreasing.
In the exercise, this principle is used to verify if the quotient \( \frac{f}{g} \) is increasing, decreasing, or constant. For example, in part (a), \( f(x) = e^x \) and \( g(x) = e^{2x} \) are tested to check that \( \frac{f}{g} = e^{-x} \), a decreasing function as its derivative is negative.
Understanding these underlying principles provides a strong basis for analyzing various mathematical functions and predicting their behaviors across given intervals.

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