How do you find local maximum value of f using the first and second derivative tests: #f(x) = e^x(x^2+2x+1)#?

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Answer 1 · 2017-06-21T07:41:34

#f_max = f(-3) = 4e^(-3) approx 0.19914827347#

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

For a local maximum or minimum of #f(x), f'(x) =0#

Applying the product rule and the standard differential of #e^x#

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

#= e^x(x^2+4x+3)#

Setting #f'(x) =0#

#e^x(x^2+4x+3) =0#

Since #e^x >0 forall x#

#x^2+4x+3 =0#

#(x+3)(x+1) = 0 -> x= -3 or -1#

To test for a maximum or minimum we need to test the sign of #f''(x)# at the turning points.

Applying the product rule and the standard differential of #e^x# again.

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

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

#f''(-3) = e^(-3)(9-18+7) = -2e^(-3) <0 -> f(-3)# is a local maximum

#f''(-1) =e^(-1)(1-6+7) = 2e^(-1)>0 -> f(-1)# is a local minimum

We are asked to find the local maximum.

#f_"max" = f(-3) = e^(-3)(9-6+1) = 4e^(-3) approx 0.19914827347#

The local extrema can be seen on the graph of #f(x)# below.
graph{e^x(x^2+2x+1) [-4.783, 0.086, -1.156, 1.277]}