Question

Va rog...................

Answer 1

Calculam f'(x).

f'(x)=[(ln²x)'*x-ln²x*x']/x²=(2*lnx*1/x*x-ln²x)/x²=(2lnx-ln²x)/x².

Rezolvam ecuatia f'(x)=0.

O fractie e 0, cand numaratorul este 0:

f'(x)=0

2lnx-ln²x=0

Notam lnx=t.

2t-t²=0

t(2-t)=0

t=0

2-t=0 => t=2

Caz 1: lnx=0 => x=1 - nu este solutie deoarece x>1 (si nu x≥1)

Caz 2: lnx=2 => x=e² - solutie

Facem tabel de semn:

x |1________e_________e²_________e³_______infinit

f' |+++++++++++++++++++++0----------------------------------------------

f |__ crescatoare______f(e²)____descrescatoare________

f'(e)=(2lne-ln²e)/e²=(2-1)/e²=1/e²>0

f'(e³)=(2lne³-ln²e³)/e⁶=(6-3*3)/e⁶=-3/e⁶<0

f(e²)=ln²e²/e²=2*2/e²=4/e²

Punctul avand abscisa e² si ordonata 4/e² este punct de maxim.

f(x)≤f(e²)

ln²x/x≤4/e²

e²*ln²x≤4x

(elnx)²≤(2√x)²

elnx≤2√x, ∀x>1