Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - P.S. Problem Solving - Page 505: 4b

Answer

False

Work Step by Step

$f(t)$ and $f(\frac{1}{2}t)$ have different peiods: $c$ and $2c$. So, the property $f(t+c)=f(t)$ can not be applied. Also, $f(t+\frac{1}{2}c)=f(t-\frac{1}{2}c)\ne f(\frac{1}{2}t)$
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