College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 2, Functions - Chapter 2 Review - Exercises - Page 270: 92


Yes, one-to-one.

Work Step by Step

We are given: $r(x)=2+\sqrt{x+3}$ This function is one-to-one because it passes the horizontal line test (it is a shifted square root function). We can show this algebraically. We start with the assumption: $x_1 \neq x_2$ $x_1+3\neq x_2+3$ $\sqrt{x_1+3}\neq \sqrt{x_2+3}$ $2+\sqrt{x_1+3}\neq 2+\sqrt{x_2+3}$ Since square rooting a unique value produces a unique value. Thus the function will not have the same $y$ value for two different $x$ values. It is therefore one-to-one.
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