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 643: 59

Answer

$y=2.1213\left(1.4142^{x}\right)$

Work Step by Step

The exponential function we want has the form $y=Ab^{x}$. $\left[\begin{array}{ll} \text{point on} & \text{corresponding}\\ \text{the graph} & \text{equation}\\ (1,3) & 3=Ab^{1}\\ (3,6) & 6=Ab^{3} \end{array}\right]$ Dividing the equations, $\displaystyle \frac{6}{3}=\frac{Ab^{3}}{Ab^{1}}$ $2=b^{2}$, so $b=\sqrt{2}\approx 1.4142$ Back-substituting into the second equation, $3=A(1.4142)^{1}$ $A=\displaystyle \frac{3}{1.4142}\approx 2.1213$ The model is $y=Ab^{x}=2.1213\left(1.4142^{x}\right)$
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