Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 1 - Linear Functions and Change - Exercises and Problems for Section 1.1 - Skill Refresher: S8

Answer

The answer is $-12$.

Work Step by Step

We insert $x = -\frac{3}{4}$ in $\frac{4}{1 + \frac{1}{x}}$: $\frac{4}{1 + \frac{1}{x}} = \frac{4}{1 + \frac{1}{-\frac{3}{4}}} = \frac{4}{1 - \frac{1}{\frac{3}{4}}} = \frac{4}{1 - (1 \div \frac{3}{4})} = \frac{4}{1 - (1 \cdot \frac{4}{3})} = \frac{4}{1 - \frac{4}{3}} = \frac{4}{\frac{3}{3} - \frac{4}{3}} = \frac{4}{ \frac{3 - 4}{3}} = \frac{4}{ \frac{- 1}{3}} = - \frac{4}{ \frac{ 1}{3}} = - 4 \div \frac{1}{3} = -4 \cdot 3 = -12$.
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