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: 9


This function is one-to-one.

Work Step by Step

This function is one-to-one. Let $x_1\neq x_2$. Then $f(x_1)=2x_1-3$ and $f(x_2)=2x_2-3$. Subtracting, we get $$f(x_1)-f(x_2)=2x_1-3-(2x_2-3)=2(x_1-x_2)$$ since $x_1\neq x_2$ then $x_1-x_2\neq 0$ so we have that $f(x_1)-f(x_2)\neq0$ i.e. $f(x_1)\neq f(x_2)$. All of this means that for every $x_1\neq x_2$ also $f(x_1)\neq f(x_2)$ so the function is one-to-one.
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