University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 50: 71


$g(x)=-f(x)$ is an one-to-one function.

Work Step by Step

If $f(x)$ is one to one, according to defintition, that means $$f(x_1)\ne f (x_2)\hspace{1cm}\text{whenever}\hspace{1cm}x_1\ne x_2$$ So we can deduce that $$-f(x_1)\ne -f(x_2)\hspace{1cm}\text{whenever}\hspace{1cm}x_1\ne x_2$$ Now consider $g(x)$, since $g(x)=-f(x)$, that means $$g(x_1)=-f(x_1)\hspace{1cm}\text{and}\hspace{1cm}g(x_2)=-f(x_2)$$ However, as we proved above $-f(x_1)\ne -f(x_2)$, hence: $$g(x_1)\ne g(x_2)\hspace{1cm}\text{whenever}\hspace{1cm}x_1\ne x_2$$ According to the definition of one-to-one functions, $g(x)$ is an one-to-one function.
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