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

1 Answer
Oct 19, 2016

#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.