Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise - Page 463: 17



Work Step by Step

The given curve in the question passing through the points (1, 6) and (3, 24). Therefore, $f(1)=Cb^{1}=6$ $f(3)=Cb^{3}=24$ Further, $\frac{f(3)}{f(1)}=\frac{Cb^{3}}{Cb^{1}}=\frac{24}{6}$ This implies $b^{2}=4$ $b=2$ Also, $Cb^{1}=6$ Thus, $C=\frac{6}{b}$ Plug 2 for $b$. $C=\frac{6}{2}=3$ Hence, the required exponential function is $f(x)=3(2^{x})$.
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