Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.1 Real Numbers, Functions, and Graphs - Exercises - Page 10: 23

Answer

The match is as follow: (a) - (i) (b) - (iii) (c) - (v) (d) - (vi) (e) - (ii) (f) - (iv)

Work Step by Step

(a) a > 3, this implies that the value of a is greater than 3. So, a simply lies on the right side of 3 in the graph. Hence,option (i). (b) |a-5| < 1/3 modules of 'a-5' represents the positive value and in graphs we read it as distance between 'a' and 5. Therefore, this statement means the distance between 'a' and 5 is less than 1/3. Hence, option (iii) matches with explanation. (c) |a - 1/3| < 5. As from the part (b), modules of 'a' and 1/3 represents the distance between 'a' and 1/3 which is less than 5 units. So, option (v) matches here. (d) |a| > 5, removing the modules: a > 5 or a < -5. From here we can see 'a' is greater than 5 and less than -5. So, 'a' either lies on the right of 5 or left side of the -5 in the graph. Hence, option (vi). (e) |a - 4| < 3 removing the modules: a - 4 < 3 or a - 4 >-3 in case 1: a-4<3, taking 4 on the right side a<7 and in case 2: a-4>-3, taking 4 on the right side a>1. So, combining both will give us 1
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