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

Question

1353 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-19T16:56:50

#x = 1+sqrt5/2=2.118#, nearly.

#x>=0#, to make #sqrt x# real.

#f(x)=sqrtx sqrt(e^-x^2)=x^(1/2)e^(-x^2/2)# = 0, when x = 0.

#f'=1/2x^(-1/2)e^(-x^2/2)+x^(1/2)e^(-x^2/2)(-x)#, x > 0.

#=x^(1/2)e^(-x^2/2)(1/(2x)-x)#

#=f(1/(2x)-x)=0#, when #x = 1/sqrt 2#.

#f''=f'(1/(2x)-x)+f(-1(2x^2)-1)#, x > 0

#=(1/(2x)-x)^2sqrtx e^(-x^2/2)-sqrtx e^(-x^2/2)(1/(2x^2)+1)#

#=sqrtxe^(-x^2/2)(1/(4x^2)+x^2-1-1/(2x^2)-1)#

#=f(x^2-2-1/(4x^2))#=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]}