/*! 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} Q. 38 In Problems 29 – 44, for the g... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Problems 29 – 44, for the given functions f and g, find:

(a)f∘g

(b)g∘f

(c)f∘f

(d) g∘g

State the domain of each composite function.

f(x)=xx+3g(x)=2x

Short Answer

Expert verified

a)f∘g=23x+2and its domain is:x|x≠0,x≠-23

b) g∘f=2(x+3)x, and its domain:x|x≠0,x≠-3

c)f∘f=x4x+9 and its domain is:x|x≠-3,x≠-94.

d)g∘g=x and its domain is:x|x≠0

Step by step solution

01

Step 1. Given information:

The functions are:

f(x)=xx+3g(x)=2x

The domain of f(x)={x|x≠-3}

The domain ofg(x)={x|x≠0}

02

Part (a)  Step 1. To find f∘g and its domain: 

f∘g=f(g(x))=2x2x+3=2x×x2+3x=23x+2

This is defined when g(x)is defined and 3x+2≠0x≠-23

So the domain is:x|x≠-23,x≠0

03

Part (b)  Step 1. To find g∘fand its domain: 

g∘f=g(f(x))=2xx+3=2(x+3)x

This is defined when f(x)is defined andx≠0.

So domain is:localid="1648202205530" x|x≠-3,x≠0

04

Part (c)  Step 1. To find  and its domain: 

f∘f=f(f(x))=xx+3xx+3+3=xx+34x+9x+3=x4x+9

This is only defined when f(x)is defined and 4x+9≠0x≠-94

So domain is:x|x≠-3,x≠-49

05

Part (d)  Step 1. To find g∘g and its domain: 

g∘g=22x=x

This is defined when g(x)is defined.

So, domain is:x|x≠0

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.