How do you find the point of inflexion of y=xe^(-x)+3? Then the inflectional tangent?

Question

#y=xe^-x+3#

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Answer 1 · 2016-09-30T18:21:10

The inflection point would be #2, 2e^(-2) +3#
The tangent line would then be
#y-(2e^(-2) +3)=-e^(-2) (x-2)#

y'=#e^(-x) -xe^(-x)#

y"=#-e^(-x) -( e^(-x) -xe^(-x) )#

= #xe^(-x) -2e^(-x)#

For inflection point make y''=0. which would give x=2

The inflection point would be #2, 2e^(-2) +3#

The slope at this point would be #e^(-2) -2e^(-2)= -e^(-2)#

The tangent line would then be
#y-(2e^(-2) +3)=-e^(-2) (x-2)#