# What are the points of inflection, if any, of f(x)=sqrt(xe^(-x^2) ?

Nov 19, 2016

$x = 1 + \frac{\sqrt{5}}{2} = 2.118$, nearly.

#### Explanation:

$x \ge 0$, to make $\sqrt{x}$ real.

$f \left(x\right) = \sqrt{x} \sqrt{{e}^{-} {x}^{2}} = {x}^{\frac{1}{2}} {e}^{- {x}^{2} / 2}$ = 0, when x = 0.

$f ' = \frac{1}{2} {x}^{- \frac{1}{2}} {e}^{- {x}^{2} / 2} + {x}^{\frac{1}{2}} {e}^{- {x}^{2} / 2} \left(- x\right)$, x > 0.

$= {x}^{\frac{1}{2}} {e}^{- {x}^{2} / 2} \left(\frac{1}{2 x} - x\right)$

$= f \left(\frac{1}{2 x} - x\right) = 0$, when $x = \frac{1}{\sqrt{2}}$.

$f ' ' = f ' \left(\frac{1}{2 x} - x\right) + f \left(- 1 \left(2 {x}^{2}\right) - 1\right)$, x > 0

$= {\left(\frac{1}{2 x} - x\right)}^{2} \sqrt{x} {e}^{- {x}^{2} / 2} - \sqrt{x} {e}^{- {x}^{2} / 2} \left(\frac{1}{2 {x}^{2}} + 1\right)$

$= \sqrt{x} {e}^{- {x}^{2} / 2} \left(\frac{1}{4 {x}^{2}} + {x}^{2} - 1 - \frac{1}{2 {x}^{2}} - 1\right)$

$= f \left({x}^{2} - 2 - \frac{1}{4 {x}^{2}}\right)$=0, when

#4x^4-8x^2-1=0 to x = 1+sqrt5/2=2.118, nearly

f'''=f'(x^2-2-1/(4x^2))+f(2x+1/(2x^3))=0+ positive number, at the zero

x=2.118 of f''.

Conclusion: x = 2.118, nearly, is a point of inflexion. I am fortunate.

The graph supports me, showing tangent crossing the curve, here.

graph{sqrt (x e^(-x^2) [-5, 5, -2.5, 2.5]}