Question #2e467

1 Answer
May 21, 2015

Answer: #f(x) = 3*0.2^x#

(1) #f(-1)=15 => 15 = a*b^-1 => 15 = a* 1/b => a=15*b#
(2) #f(1)=3/5 => 3/5 = a*b^1 => 3/5 = a*b #

From (1) we can substitute #a# with #15 b# in the equation (2):

#15*b^2 = 3/5 => b^2 = 3/75# #=># #b=0.2# #=> a= 3#

Our function is #f(x)=3*0.2^x#