## Calculus: Early Transcendentals 8th Edition

$f(x)=3\cdot2^x$
At $(1,6), 6=Cb^1=Cb$ $(i)$ At $(3,24), 24=Cb^3$ $(ii)$ $\frac{(ii)}{(i)}: b^2=4, b=\pm 2$ $\because$ The graph is monotically increasing, $b\gt 0,$ $\therefore b=2\rightarrow(i)$ $2C=6, C=3$ $\therefore f(x)=3\cdot2^x$