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: 29

Answer

See proof: solve the equations given by the two conditions, plug in the greatest and least values of a and b into the final condition.

Work Step by Step

Solve the equations given by the two conditions The first condition: $|a − 5| < \frac{1}{2}$ $a-5-\frac{1}{2}$ $a<5\frac{1}{2}$ and $a>4\frac{1}{2}$ The second condition: $|b − 8| < \frac{1}{2}$ $b-8-\frac{1}{2}$ $b<8\frac{1}{2}$ and $b>7\frac{1}{2}$ The hint is to use the triangle inequality $(|a + b|≤|a| +|b|)$, but I did not use the hint. To prove $|(a + b) − 13| < 1$, we take the greatest values that a and b could possibly be. $|<5\frac{1}{2}+<8\frac{1}{2}-13|$ $|<14-13|$ To prove $|(a+b)−13|<1$, we also take the least values that a and b could possibly be. $|>4\frac{1}{2}+>7\frac{1}{2}-13|$ $|>12-13|$ These value has to be $<1$.
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