Answer
$f(x)=2\cdot(\frac{2}{3})^{-x}$
Work Step by Step
$$f(x)=Cb^{-x}$$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, $C\gt0$
$\therefore C=2\rightarrow(i)$
$b=\frac{2}{3}$
$\therefore f(x)=2\cdot(\frac{2}{3})^{-x}$
