Answer:
f(x) =3/2 (8/7)^(–x) is the only expo. decay function here.
Step-by-step explanation:
First, let's write each of these functions on its own line for greater clarity:
f(x) =3/4 (7/4)x
f(x) =2/3 (4/5)–x
f(x) =3/2 (8/7)–x
f(x) =1/3 (9/2)x
Next, use " ^ " to indicate exponentiation:
f(x) =3/4 (7/4)^x
f(x) =2/3 (4/5)^(–x)
f(x) =3/2 (8/7)^(–x)
f(x) =1/3 (9/2)^x
An exponential growth function looks like f(x) = a(base)^x, where the base is greater than one.
Therefore, we eliminate f(x) =3/4 (7/4)^x; the base (7/4) is greater than 1.
The next one, f(x) =2/3 (4/5)^(–x), is equivalent to f(x) =2/3 (5/4)^x. This is a growth function because the base is 5/4 (greater than 1).
The next one, f(x) =3/2 (8/7)^(–x), is equivalent to f(x) =3/2 (7/8)^(x). Here the base is between 0 and 1, so this is the exponential decay function.
The last one has a base greater than 1, so is an expo. growth function.
