BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 1.3, Problem 35E

a.

To determine

To find:the given function fg and their domain.

Expert Solution

Answer to Problem 35E

  (fg)(x)=2x2+6x+5(x+1)(x+2);{1,2} .

Explanation of Solution

Given:

The given function is f(x)=x+1x,g(x)=x+1x+2 .

Calculation:

As the given function is f(x)=x+1x,g(x)=x+1x+2 .

  (fg)(x)=f(g(x))=x+1x+2+1x+1x+2{as f(x)=x+1x}=x+1x+2+x+2x+1{11x=x}=(x+1)2+(x+2)2(x+1)(x+2)=(x2+2x+1)+(x2+4x+4)(x+1)(x+2)=2x2+6x+5(x+1)(x+2)

Since, g(x) is not defined at x=1,2

Therefore, the domain of the function {1,2} .

Hence,the range and domain are (fg)(x)=2x2+6x+5(x+1)(x+2);{1,2}

b.

To determine

To find: the given function gf and their domain.

Expert Solution

Answer to Problem 35E

  (gf)(x)=1x3;domain is{x|x0,x} .

Explanation of Solution

Given:

The given function is f(x)=x+1x,g(x)=x+1x+2 .

Calculation:

As the given function is f(x)=x+1x,g(x)=x+1x+2 .

  (gf)(x)=g(f(x))=x+1x+1x+1x+2{as g(x)=x+1x+2}=x2+x+1x2+2x+1=x2+x+1(x+1)2{as (x+1)2=x2+2x+1}

Since, g(x) is not defined at x=0,1

Therefore, the domain of the function {0,1} .

Hence, the range and domain are (gf)(x)=x2+x+1(x+1)2;{0,1}

c.

To determine

To find: the given function ff and their domain.

Expert Solution

Answer to Problem 35E

  (ff)(x)=x4+3x2+1x(x2+1);domain is{0} .

Explanation of Solution

Given:

The given function is f(x)=x+1x,g(x)=x+1x+2 .

Calculation:

As the given function is f(x)=x+1x,g(x)=x+1x+2 .

  (ff)(x)=f(g(x))=x+1x+1x+1x{as f(x)=x+1x}=x2+1x+1x2+1x=x2+1x+xx2+1{11x=x}=(x2+1)2+x2(x2+1)x=(x4+2x2+1)+(x2)(x+1)x=x4+3x2+1x(x2+1)

Since, g(x) is not defined at x=0

Thus, the domain is {0}

Hence, the range and domain are (ff)(x)=x4+3x2+1x(x2+1);domain is{0}

d.

To determine

To find: the given function gg and their domain.

Expert Solution

Answer to Problem 35E

  (gg)(x)=2x+33x+5;{2,53} .

Explanation of Solution

Given:

The given function is f(x)=x+1x,g(x)=x+1x+2 .

Calculation:

As the given function is f(x)=x+1x,g(x)=x+1x+2 .

  (gf)(x)=g(f(x))=x+1x+2+1x+1x+2+2{as g(x)=x+1x+2}=x+1+x+2x+2x+1+2(x+2)x+2=x+1+x+2x+1+2x+4=2x+33x+5

Since, g(x) is not defined at x=2,53

Therefore, the domain of the function {2,53} .

Hence, the range and domain are (gg)(x)=2x+33x+5;{2,53}

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