Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises: 18



Work Step by Step

The given curve in the question passing through the points $(0,2)$ and $(2,\frac{2}{9})$. Therefore, $f(0)=Cb^{0}=2$ $f(2)=Cb^{2}=\frac{2}{9}$ Further, $\frac{f(2)}{f(0)}=\frac{Cb^{2}}{Cb}=\frac{2/9}{2}$ This implies $b^{2}=\frac{1}{9}$ $b=\frac{1}{3}$ Also, $Cb^{2}=\frac{2}{9}$ Thus, $C=\frac{2/9}{b}$ Plug $\frac{1}{3}$ for $b$. $C=2$ Hence, the required exponential function is $f(x)=2(\frac{1}{3})^{x}$.
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