Answer:
f(x) = [tex](\frac {1}{2})^{(6x)}[/tex]
Step-by-step explanation:
The given options are,
a) f(x) = [tex]2^{(6x)}[/tex]
b) f(x) = [tex](\frac {1}{2})^{(6x)}[/tex]
c) f(x) = [tex]2 \times (\frac {1}{6})^{x}[/tex]
d) f(x) = [tex]\frac {1}{2} \times (\frac {1}{6})^{x}[/tex]
Now, clearly a) is a monotonically increasing function, hence discarded, and both of c) and d) don't pass through (0, 1) hence they are also discarded.
Only b) is a decay function which does also pass through (0, 1), hence, b) is the correct option.
