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

Nov 13, 2016

$f \left(x\right)$ is concave at $x = 1$.

#### Explanation:

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

$f \left(x\right) = {x}^{3} - x {e}^{- {x}^{2} + x}$

$f ' \left(x\right) = 3 {x}^{2} - \left[\left(x\right) \left(- 2 x + 1\right) \left({e}^{- {x}^{2} + x}\right) + \left({e}^{- {x}^{2} + x}\right)\right]$
$f ' \left(x\right) = 3 {x}^{2} - \left(x\right) \left(- 2 x + 1\right) \left({e}^{- {x}^{2} + x}\right) - \left({e}^{- {x}^{2} + x}\right)$
$f ' \left(x\right) = 3 {x}^{2} - \left(- 2 {x}^{2} + x\right) \left({e}^{- {x}^{2} + x}\right) - \left({e}^{- {x}^{2} + x}\right)$

$f ' ' \left(x\right) = 6 x - \left[\left(- 2 {x}^{2} + x\right) \left(- 2 x + 1\right) \left({e}^{- {x}^{2} + x}\right) + \left(- 4 x + 1\right) \left({e}^{- {x}^{2} + x}\right)\right] - \left(- 2 x + 1\right) \left({e}^{- {x}^{2} + x}\right)$

$f ' ' \left(x\right) = 6 x - \left(- 2 {x}^{2} + x\right) \left(- 2 x + 1\right) \left({e}^{- {x}^{2} + x}\right) - \left(- 4 x + 1\right) \left({e}^{- {x}^{2} + x}\right) - \left(- 2 x + 1\right) \left({e}^{- {x}^{2} + x}\right)$

It's hard to find zeros of the ugly equation of $f ' ' \left(x\right)$, 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 $f \left(x\right)$ is concave from $\approx \left(.186 , \infty\right)$, and $f \left(x\right)$ is convex from $\approx \left(- \infty , .186\right)$

Therefore, $f \left(x\right)$ is concave at $x = 1$.