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: 22

Answer

Both are exponential. $f(x)=2.7\cdot 3^{x},\displaystyle \quad g(x)=0.75\cdot(\frac{1}{2})^{x}$

Work Step by Step

If a function x is an exponential function, then it has form $f(x)=a\cdot b^{x}$ When we observe an increase of x by 1 unit, $f(x+1)=ab^{x+1}.$ what follows is that the ratio $\displaystyle \frac{f(x+1)}{f(x)}=\frac{ab^{x+1}}{ab^{x}}=b.$ (the ratio is constant when arguments increase by 1). Thus, if the x increases by unit, the function value is multiplied with b. We use this to recognize whether a table of function values represent an exponential function or not. --- The x-values increase by 1 unit. For the values of f(x) we observe $ \displaystyle \frac{f(x+1)}{f(x)}$: $\displaystyle \frac{0.9}{0.3}=3,\quad\frac{2.7}{0.9}=3,\quad\frac{8.1}{2.7}=3,\quad\frac{24.3}{8.1}=3,\quad$ so f is exponential, $f(x)=a\cdot 3^{x}$ Since $f(0)=2.7$, then $a=2.7$ Thus, $f(x)=2.7\cdot 3^{x}$ For the values of g(x), $\displaystyle \frac{1.5}{3}=\frac{1}{2},\quad\frac{0.75}{1.5}=\frac{1}{2},\quad\frac{0.375}{0.75}=\frac{1}{2},\quad\frac{0.1875}{0.375}=\frac{1}{2}$ so it is exponential, $g(x)=a\displaystyle \cdot(\frac{1}{2})^{x}.$ Since $g(0)=0.75$, then $a=0.75$ Thus, $g(x)=0.75\displaystyle \cdot(\frac{1}{2})^{x},\qquad ($which can also be written as $\displaystyle \frac{3}{4}\cdot 2^{-x})$
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