f(x) = e^x(x^2+2x+1)f(x)=ex(x2+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
