#f(x)=x^2e^(11−x)#
#f'(x)=2xe^(11−x)+x^2e^(11−x)*-1#
#f'(x)=e^(11−x)[2x-x^2]#
#f''(x)=-e^(11−x)(2x-x^2)+e^(11−x)(2-2x)#
#f''(x)=e^(11−x)[-2x+x^2+2-2x]#
#f''(x)=e^(11−x)[x^2-4x+2]#
#e^(11−x)>0quad# That means we don't care about that. BUT:
#x^2-4x+2=0#
#x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)#
#x_(1,2)=(4+-sqrt(4^2-4*1*2))/(2*1)#
#x_(1,2)=(4+-sqrt(16-8))/(2)=(2*2+-sqrt(2^2*2))/(2)#
#x_(1,2)=(cancel2*2+-cancel2sqrt(2))/(cancel2)=2+-sqrt(2)#
#x_1=2-sqrt(2)#
#x_2=2+sqrt(2)#