Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.1 Functions - Exercises Set 0.1: 7

Answer

(a) f(0) = -2 f(2) = 10 f(-2) = 10 f(3) = 25 f($\sqrt 2$) = 4 f(3t) = 27t² - 2 (b) f(0) = 0 f(2) = 4 f(-2) = -4 f(3) = 6 f($\sqrt 2$) = 2$\sqrt 2$ f(3t) = 1/3t, if t $\gt$ 1. f(3t) = 2(3t) = 6t, if t $\leq$ 1

Work Step by Step

(a) f(0) = 3(0)² - 2 = -2 f(2) = 3(2)² - 2 = 3 $\times$ 4 - 2 = 10 f(-2) = 3(-2)² - 2 = 3 $\times$ 4 - 2 = 10 f(3) = 3(3)² - 2 = 3 $\times$ 9 - 2 = 25 f($\sqrt 2$) = 3($\sqrt2$)² - 2 = 3 $\times$ 2 - 2 = 4 f(3t) = 3(3t)² - 2 = 3 $\times$ 9t² - 2 = 27t² - 2 (b) f(0) = 2 $\times$ 0 = 0 f(2) = 2 $\times$ 2 = 4 f(-2) = 2 $\times$ (-2) = -4 f(3) = 2 $\times$ 3 = 6 f($\sqrt 2$) = 2$\sqrt 2$ 3t = 3 -> t = 1, so f(3t) = 1/3t, if t $\gt$ 1. f(3t) = 2(3t) = 6t, if t $\leq$ 1
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