How do you find points of inflection and determine the intervals of concavity given #y=xe^x#?

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Answer 1 · 2017-02-15T14:33:57

Point of inflection is at #x=-2# and while in range #(-oo,-2)# the curve is concave down and in range #(-2,oo)# the curve is concave up.

At points of inflection, second derivative of the function is equal to zero. Hence le us first wok out second derivative for #y=xe^x#.

As #y=xe^x# and

#(dy)/(dx)=x xx d/(dx)e^x+1xxe^x=xe^x+e^x=e^x(x+1)#

and #d^2/(dx)^2e^x(x+1)#

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

and this is zero at #x=-2#

As there is only one point of inflection, we have two ranges #(-oo,-2)# and (-2,oo)#

In the range #(-oo,-2)#, second derivative is negative and hence in this range the curve is concave down and

in the range #(-2,oo)#, second derivative is positive and hence in this range the curve is concave up.
graph{xe^x [-7.4, 2.6, -1.56, 3.44]}