What are the global and local extrema of #f(x) = e^x(x^2+2x+1) # ?

1 Answer
Feb 28, 2017

Answer:

#f(x)# has an absolute minimum at #(-1. 0)#
#f(x)# has a local maximum at #(-3, 4e^-3)#

Explanation:

#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#