We have to find 1st and 2nd derivative:
f'(x)=(1/x*8sqrtx-8/(2sqrtx)lnx)/(64x)=(1/sqrtx-lnx/(2sqrtx))/(8x)=
=(2-lnx)/(16xsqrtx)
f^((2))(x)=(-1/x*16xsqrtx-16*3/2*sqrtx*(2-lnx))/(16^2x^3)=
=(-16sqrtx-24sqrtx(2-lnx))/(16^2x^3)=
=(-8sqrtx(2+6-3lnx))/(16^2x^3)=(-sqrtx(8-3lnx))/(32x^3)
Stationary points are zeros of 1st derivative:
f'(x_s)=0 <=> (2-lnx_s)/(16x_ssqrtx_s)=0 <=>
(2-lnx_s=0 ^^ 16x_ssqrtx_s!=0) <=>
(lnx_s=2 ^^ x_s!=0) <=> (x_s=e^2 ^^ x_s!=0) <=> x_s=e^2
AAx>x_s: lnx>2 ^^ 2-lnx<0 ^^ 16xsqrtx>0
AAx>x_s : f'(x)<0, f is decreasing
AAx< x_s ^^ x>0 : lnx<2 ^^ 2-lnx>0 ^^ 16xsqrtx>0
AAx< x_s ^^ x>0 : f'(x)>0, f is increasing
For x=x_s function f has maximum value:
f_max=f(e^2)=lne^2/(8sqrt(e^2))=2/(8e)=1/(4e)
Inflection points are zeros of 2nd derivative if 2nd derivative changes sign in those points and iff function is continuous :
f^((2))=0 <=> (-sqrtx_i(8-3lnx_i))/(32x_i^3)=0 <=>
(-sqrtx_i(8-3lnx_i)=0 ^^ 32x_i^3!=0) <=>
(8-3lnx_i=0 ^^ x_i!=0) <=> (lnx_i=8/3 ^^ x_i!=0) <=>
(x_i=e^(8/3) ^^ x_i!=0) <=> x_i=e^(8/3)=e^2root(3)(e^2)
So, x_i=e^2root(3)(e^2) is potential inflection point.
AAx>x_i: 8-3lnx<0, f^((2))>0
AAx< x_i ^^ x>0: 8-3lnx>0, f^((2))<0
Indeed, x_i is inflection point:
f(x_i)=f(e^(8/3))=(lne^(8/3))/(8sqrt(e^(8/3)))=(8/3)/(8e^(4/3))=1/(3eroot(3)(e))
Note: f is defined for AAx>0 because of lnx, so I used that fact.