a)f `(x)=4x³/4-1/x=x³-1/x
b) Rezolvi ecuatia f `(x)=0
x³-1/x=0
(x^4-1)/x=0
(x²-1)(x²+1)/x=0 x²+1>0∀x>0
x²-1=0
x1=-1<0 nu se accepta
x2=1 solutie
Calculezi semnul lui f `(x) de-o parte si de alta a lui 1
Pt x∈(0,1) F `(X)<0
x>1 f `(x)>0 => x=1 punct de extrem (minim)
c)fie g(x)=ln√x-(x²-1)/4 g:(0,∞)→R
g `(x)=1/2√x)*1/√x-2x/4=1/2x-x/2=(1-x²)/2x
g `(x)=(1-x²)/2x=0 x1=-1<0 nu se accepta, x2=1
pt x∈(0,1) g `(x)>0 si pt x>1 g `(x)<0 =>x=1 punct de maxim
Adica g(1)=0 >g(x) ∀x>0=> g(x)<0
ln√x-(x²-1)/4<0=>
ln√x<(x²-1)/4 concluzia
