Find the point of inflection of the curve, #y=xe^x, x in RR#, if any. What is it's radius of curvature at #x = 2#?

1 Answer
Jim H
Jan 3, 2018

It depends on your definition of inflection point.

Explanation:

If your definition requires a horizontal tangent line at an inflection point then the curve has no inflection points.

#y' = e^x(x+1)# which is #0# only at #x=-1#. But the curve has a minimum at #x=-1# not an inflection points.

If you define an inflection points as a point on the curve at which the concavity changes then the point #(-2,-2e^-2)# is a point of inflection.

#y'' = e^x(x+2)# which is #0# at #x=-2# and #y'' 0# for #x < -2# while #y'' > 0# for #x > -2#.

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