Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.1 Inverse Functions - 6.1 Exercises - Page 406: 13


This is not a one-to-one function.

Work Step by Step

Let $y=h(x)=1+\cos x$. $y=1$ for every $x=(2k-1)\frac{\pi}{2}$, where $k$ is an integer, because $\cos (2k-1)\frac{\pi}{2}=0$ (cosine of the odd multiple of the right angle is zero). So to multiple values of the argument $x$ corresponds the same value $y=0$ i.e. the function is not one-to-one.
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