#f(x) = x*3^(-x)#
#f'(x) = x* d/dx 3^(-x) + 3^(-x) *1# [Product rule]
# d/dx 3^(-x) = -ln3*3^(-x)#
#:. f'(x) = x*( -ln3*3^(-x)) + 3^(-x)#
#= -3^(-x)(xln3-1)#
For critical points of #f(x), f'(x) =0#
Since #-3^(-x) != 0 forall x in RR#
#xln3-1=0#
#x= 1/ln3 approx 0.910#
#:. f(approx 0.910)# will be a critical value of #f(x)#
To determine whether the critical point is a local maximum or minimum.
#f''(x) = -3^(-x)*ln3 +(xln3-1)*ln3*3^(-x)#
#= -3^(-x)ln3(1-xln3+1)#
#= -3^(-x)ln3(2-xln3)#
Consider: # -3^(-x)ln3 < 0 forall x in RR#
And: #2- (~=0.910)*ln3 = (2- ~=1.00) >0 #
#:. f''(~=0.910)<0 -> f(~=0.910)# is a local maximum
#f_max = f(~=0.910) approx 0.335#
This can be seen on the graph of #f(x)# below.
graph{ x*3^(-x) [-1.415, 5.515, -1.498, 1.966]}