   Chapter 1.1, Problem 25E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# If f(x) = 3x2 ‒ x + 2, find f(2), f(‒2), f(a), f(‒a), f(a + 1), 2f(a), f(2a), f(a2), [f(a)]2, and f(a + h).

To determine

To find: The values of f(2),f(2),f(a),f(a),f(a+1),2f(a),f(2a), f(a2),[f(a)]2and f(a+h) .

Explanation

Given:

The function f(x)=3x2x+2 .

Calculation:

Evaluate f(2) by replacing x by 2 in the expression for f(x) .

f(2)=3(2)22+2=3(4)=12

Therefore, the value of f(2)=12 .

Evaluate f(2) by replacing x by −2 in the expression for f(x) .

f(2)=3(2)2(2)+2=3(4)+2+2=12+4=16

Therefore, the value of f(2)=16 .

Evaluate f(a) by replacing x by a in the expression for f(x) .

f(a)=3(a)2a+2=3a2a+2

Therefore, the value of f(a)=3a2a+2 .

Evaluate f(a) by replacing x by −a in the expression for f(x) .

f(a)=3(a)2(a)+2=3a2+a+2

Therefore, the value of f(a)=3a2+a+2 .

Evaluate f(a+1) by replacing x by a+1 in the expression for f(x) .

f(a+1)=3(a+1)2(a+1)+2=3(a2+2a+1)a1+2=3a2+6a+3a1+2=3a2+5a+4

Therefore, the value of f(a+1)=3a2+5a+4 .

Evaluate 2f(a) by multiplying f(a) by 2 in the expression for f(a) .

2f(a)=2f(a)=2(3a2a+2)            [Since f(a)=3a2a+2]=6a22a+4

Therefore, the value of 2f(a)=6a22a+4

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