Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.3 Exponential Functions - 5.3 Assess Your Understanding - Page 281: 34


Exponential $F(x)=\left(1/2\right)^{x+2}$

Work Step by Step

The ratio of consecutive values is constant $\left(\text{because }\dfrac{\frac{1}{4}}{\frac{1}{2}}=\dfrac{\frac{1}{8}}{\frac{1}{4}}=\dfrac{\frac{1}{16}}{\frac{1}{8}}=\dfrac{\frac{1}{32}}{\frac{1}{16}}=\dfrac{1}{2}\right)$ hence, the function is exponential with a common ratio of $\dfrac{1}{2}$. The difference of consecutive values is not constant ($\frac{1}{4}-\frac{1}{8}\ne \frac{1}{2}-\frac{1}{4}$) hence, the function is not linear. A function that models the data: (with a common ratio of $\frac{1}{2}$ and $F(0)=\left(\frac{1}{2}\right)^2$ is $F(x)=\left(1/2\right)^{x+2}$.
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