## Calculus 8th Edition

$f(x)=2(\frac{1}{3})^{x}$
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 $f(x)=2(\frac{1}{3})^{x}$.