#f(x) = e^x(x^2+2x+1)#
#f'(x) = e^x(2x+2) + e^x(x^2+2x+1)# [Product rule]
#= e^x(x^2+4x+3)#
For absolute or local extrema: #f'(x)= 0#
That is where: #e^x(x^2+4x+3) =0#
Since #e^x >0 forall x in RR#
#x^2+4x+3 =0#
#(x+3)(x-1) =0 -> x=-3 or -1#
#f''(x) = e^x(2x+4) + e^x(x^2+4x+3)# [Product rule]
#= e^x(x^2+6x+7)#
Again, since #e^x>0# we need only test the sign of #(x^2+6x+7)#
at our extrema points to determine whether the point is a maximum or a minimum.
#f''(-1) = e^-1 * 2 > 0 -> f(-1)# is a minimum
#f''(-3) = e^-3 * (-2) < 0 -> f(-3)# is a maximum
Considering the graph of #f(x)# below it is clear that #f(-3)# is a local maximum and #f(-1)# is an absolute minimum.
graph{e^x(x^2+2x+1) [-5.788, 2.005, -0.658, 3.24]}
Finally, evaluating the extrema points:
#f(-1) = e^-1(1-2+1) = 0#
and
#f(-3) = e^-3(9-6+1) = 4e^-3 ~= 0.199#
