## Calculus: Early Transcendentals 8th Edition

$f(x)=2\cdot(\frac{3}{2})^{-x}$
$$f(x)=Cb^{-x}$$since the graph is monotonically decreasing. At $(-1,3), 3=Cb^{1}$ $(i)$ At $(1,\frac{4}{3}), \frac{4}{3}=Cb^{-1} (ii)$ $(i)\times(ii): 4=C^2, C=\pm2$ $\because$ The graph is always positive, $C\gt0$ $\therefore C=2\rightarrow(i)$ $b=\frac{3}{2}$ $\therefore f(x)=2\cdot(\frac{3}{2})^{-x}$