Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 9 - Section 9.2 - Exponential Functions and Models - Exercises - Page 642: 11

Answer

$\begin{array}{|c|r|r|r|r|r|r|r|} \hline x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline s(x) &\displaystyle \frac{1}{16} &\displaystyle \frac{1}{8} & \displaystyle \frac{1}{4} & \displaystyle \frac{1}{2} & 1 & 2 & 4 \\ \hline \end{array}$ Technology formula:$\quad 2$^$(x-1)$

Work Step by Step

Technology formula:$\quad 2$^$(x-1)$ $f(-3)= 2^{-3-1}=2^{-4}=(\displaystyle \frac{1}{2})^{4}=\frac{1}{16}\qquad $tech:$\qquad 2$^$(-3-1)$ $f(-2)= 2^{-2-1}=2^{-3}=(\displaystyle \frac{1}{2})^{3}=\frac{1}{8}\qquad $tech:$\qquad 2$^$(-2-1)$ $f(-1)= 2^{-1-1}=2^{-2}=(\displaystyle \frac{1}{2})^{2}=\frac{1}{4}\qquad $tech:$\qquad 2$^$(-1-1)$ $f(0)= 2^{0-1}=2^{-1}=\displaystyle \frac{1}{2}\qquad $tech:$\qquad 2$^$(0-1)$ $f(1)= 2^{1-1}=2^{0}=1\qquad $tech:$\qquad 2$^$(1-1)$ $f(2)= 2^{2-1}=2^{1}=2\qquad $tech:$\qquad 2$^$(2-1)$ $f(3)= 2^{3-1}=2^{2}=4\qquad $tech:$\qquad 2$^$(2-1)$
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