Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Statement
Find \( f(x) \) and \( g(x) \) such that \( h(x) = (f \circ g)(x) \). Let \( g(x) = 3x + 1 \).
\[ h(x) = \frac{1}{\sqrt{3x + 1}} \]
\[ f(x) = \]
### Explanation of the Problem
The goal is to decompose the function \( h(x) \) into two functions, \( f(x) \) and \( g(x) \), such that the composition of \( f \) and \( g \) yields \( h \). Specifically, given \( g(x) = 3x + 1 \), we need to find the function \( f \) such that \( h(x) = f(g(x)) \).
### Decomposition Process
Given:
\[ g(x) = 3x + 1 \]
We can substitute \( g(x) \) into \( h(x) \):
\[ h(x) = \frac{1}{\sqrt{g(x)}} \]
Substituting \( g(x) = 3x + 1 \):
\[ h(x) = \frac{1}{\sqrt{3x + 1}} = \frac{1}{\sqrt{g(x)}} \]
This shows that:
\[ f(u) = \frac{1}{\sqrt{u}} \]
where \( u = g(x) \).
Thus, we have:
\[ f(x) = \frac{1}{\sqrt{x}} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a1263ae-327d-4428-a0aa-137e6771516c%2F4f99e170-aeb8-4a2b-8eb4-9bb45f0c7a26%2Fyl97g0d_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find \( f(x) \) and \( g(x) \) such that \( h(x) = (f \circ g)(x) \). Let \( g(x) = 3x + 1 \).
\[ h(x) = \frac{1}{\sqrt{3x + 1}} \]
\[ f(x) = \]
### Explanation of the Problem
The goal is to decompose the function \( h(x) \) into two functions, \( f(x) \) and \( g(x) \), such that the composition of \( f \) and \( g \) yields \( h \). Specifically, given \( g(x) = 3x + 1 \), we need to find the function \( f \) such that \( h(x) = f(g(x)) \).
### Decomposition Process
Given:
\[ g(x) = 3x + 1 \]
We can substitute \( g(x) \) into \( h(x) \):
\[ h(x) = \frac{1}{\sqrt{g(x)}} \]
Substituting \( g(x) = 3x + 1 \):
\[ h(x) = \frac{1}{\sqrt{3x + 1}} = \frac{1}{\sqrt{g(x)}} \]
This shows that:
\[ f(u) = \frac{1}{\sqrt{u}} \]
where \( u = g(x) \).
Thus, we have:
\[ f(x) = \frac{1}{\sqrt{x}} \]
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