tg π/4=1
Calculezi f `(x) si rezolvi ecuatia f `(x)=1
f `(x)=√(x-1)+(x+2)/2√(x-1)=1
[2√(x-1)²+(x+2)]/2√(x-1)=1
[2(x-1)+x+2]/2√(x-1)=1
3x=2√(x-1)
9x²=4(x-1)
9x²-4x+4=0
Determinantul Δ=-128<0.Ecuatia nu are solutii reale, deci nu exista nici o tangenta la Gf care determina < de π/4