Is #f(x)=x^3-xe^(x-x^2) # concave or convex at #x=1 #?

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Answer 1 · 2016-11-13T17:40:53

#f(x)# is concave at #x=1#.

To find intervals of convex and concave, we have to find the 2nd derivative.

#f(x)=x^3-xe^(-x^2+x)#

#f'(x)=3x^2-[(x)(-2x+1)(e^(-x^2+x))+(e^(-x^2+x))]#
#f'(x)=3x^2-(x)(-2x+1)(e^(-x^2+x))-(e^(-x^2+x))#
#f'(x)=3x^2-(-2x^2+x)(e^(-x^2+x))-(e^(-x^2+x))#

#f''(x)=6x-[(-2x^2+x)(-2x+1)(e^(-x^2+x))+(-4x+1)(e^(-x^2+x))]-(-2x+1)(e^(-x^2+x))#

#f''(x)=6x-(-2x^2+x)(-2x+1)(e^(-x^2+x))-(-4x+1)(e^(-x^2+x))-(-2x+1)(e^(-x^2+x))#

It's hard to find zeros of the ugly equation of #f''(x)#, so I graphed it:

graph{6x-(-2x^2+x)(-2x+1)(e^(-x^2+x))-(-4x+1)(e^(-x^2+x))-(-2x+1)(e^(-x^2+x)) [-10, 10, -5, 5]}

This shows that #f(x)# is concave from #approx(.186,oo)#, and #f(x)# is convex from #approx(-oo,.186)#

Therefore, #f(x)# is concave at #x=1#.