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 61E

(a)

To determine

To find: The function f(x) where (fg)(x)=h(x) .

Expert Solution

Answer to Problem 61E

The function f(x)=x2+6 .

Explanation of Solution

The given functions are g(x)=2x+1,h(x)=4x2+4x+7 .

The composite function (fg)(x)=h(x) is defined as follows.

(fg)(x)=h(x)f(g(x))=h(x)

Substitute the expression for g(x) and h(x) .

Thus, f(2x+1)=4x2+4x+7 .

It can be expressed as follows.

f(2x+1)=(2x)2+2(2x)(1)+1+6=(2x+1)2+6

Therefore, f(2x+1)=(2x+1)2+6 (1)

Replace x by 2x + 1 in equation (1) and get the expression for f(x) .

f(x)=x2+6

Thus, the function f(x)=x2+6 .

(b)

To determine

To find: The function g(x) where (fg)(x)=h(x) .

Expert Solution

Answer to Problem 61E

The function g(x)=x2+x1 .

Explanation of Solution

The given functions are f(x)=3x+5,h(x)=3x2+3x+2 .

The composite function (fg)(x)=h(x) is defined as follows.

(fg)(x)=h(x)f(g(x))=h(x)

Substitute the expression for f(x) and h(x) .

Thus, 3g(x)+5=3x2+3x+2 (2)

Notice that the degree of f(x) is 1 and the degree of h(x) 2.

Therefore, the degree of g(x) must be 2 so that the composite function f(g(x))=h(x) will be true.

Thus, let g(x)=ax2+bx+c and substitute in equation (2).

3(ax2+bx+c)+5=3x2+3x+23ax2+3bx+3c+5=3x2+3x+2

Compare the coefficients of x2 and x find the value of a and b.

3a=3a=13b=3b=1

Similarly, compare the value of constant term and find the value of c.

3c+5=23c=3c=1

Thus, the value of a=b=1,c=1 .

Therefore, g(x)=x2+x1 .

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