How do you find the local max and min for #f(x) = e^x(x^2+2x+1)#?

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Answer 1 · 2016-03-17T08:07:31

We have a local maxima at #x=-3# and a local minima at #x=-1#.

To find local maxima and minima of #f(x)=e^x(x^2+2x+1)#, let us first differentiate it

#f'(x)=e^xd/(dx)(x^2+2x+1)+(de^x)/(dx)(x^2+2x+1)# or

#f'(x)=e^x(2x+2)+e^x(x^2+2x+1)# or

#f'(x)=e^x(x^2+4x+3)=e^x(x+3)(x+1)#
which will be zero for #x=-1# and #x=-3#

Hence, maxima and minima will be at #x=-1# and #x=-3#

[For more details click here.](http://socratic.org/questions/how-do-you-find-the-local-max-and-min-for-f-x-2x-3-5x-1-in-the-interval-is-3-3#240831)

To find whether is maxima or minima, let us find out second derivative.

#f''(x)-d/(dx)e^x(x^2+4x+3)#, which is

#f''(x)=e^xd/(dx)(x^2+4x+3)+(de^x)/(dx)(x^2+4x+3)# or

#f''(x)=e^x(2x+4)+e^x(x^2+4x+3)# or

#f''(x)=e^x(x^2+6x+7)#

At #x=-3#, #f''(x)=e^-3((-3)^2+6(-3)+7)=-2e^-3# and is negative.

At #x=-1#, #f''(x)=e^-1((-1)^2+6(-1)+7)=2e^-3# and is positive.

Hence we have a local maxima at #x=-3# and a local minima at #x=-1#.

graph{e^x(x^2+2x+1) [-10, 10, -5, 5]}