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.
