Question

How do you find formulas for the exponential functions satisfying the given conditions f(3)=-3/8 and f(-2)=-12?

Answer 1

`f(x) = -3(1/2)^x`

Explanation:

Suppose we have an exponential function of the form `f(x) = Ae^(Bx)`

Given our conditions, we get the system

`{(Ae^(3B) = -3/8), (Ae^(-2B)=-12):}`

Multiplying the first equation by `e^(-3B)` and the second by `e^(2B)`, we get

`{(A = -3/8e^(-3B)), (A = -12e^(2B)):}`

Thus, equating the two, we have

`-3/8e^(-3B) = -12e^(2B)`

Multiplying by `e^(3B)/-12`, we get

`1/32 = e^(5B)`

Taking the natural log of both sides, we find

`ln(1/32) = ln(e^(5B))`

`=> ln(2^(-5)) = 5B`

`=> -5ln(2) = 5B`

`=> B = -ln(2) = ln(1/2)`

Note, then, that `e^(Bx) = (e^B)^x = (e^(ln(1/2)))^x = (1/2)^x`

Now we can substitute this back into a prior equation, say, `A=-12e^(2B)`. Doing so, we get

`A = -12(1/2)^2`

`=-12/4`

`=-3`

Thus we get our full equation as

`f(x) = Ae^(Bx) = -3(1/2)^x`

Checking we find that

`f(3) = -3(1/2)^3 = -3(1/8) = -3/8`
and
`f(-2) = -3(1/2)^(-2) = -3(4) = -12`

as desired.