Is $f (x) = x^{3} - x e^{x - x^{2}}$ $f(x)=x^3-xe^(x-x^2)$ concave or convex at $x = 1$ $x=1$ ?

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Answer 1 · 2016-11-13T17:40:53

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

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

$f (x) = x^{3} - x e^{- x^{2} + x}$ $f(x)=x^3-xe^(-x^2+x)$

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

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

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

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 $x=1$ .

Is f(x)=x^3-xe^(x-x^2) f(x)=x3−xex−x2f(x)=x^3-xe^(x-x^2) concave or convex at x=1 x=1x=1 ?