Metoda 1
Deoarece x²≥0 si x²+4≥0⇒f(x)≥0
Aratam ca f(x)<1⇔x²/(x²+4)-1<0⇔-4/(x²+4)<0(Adevarat)
Deoarece 0≤f(x)<1⇒ Imf=[0,1)
Metoda 2
Fie f(x)=y⇒x²/(x²+4)=y⇒x²=yx²+4y⇒(1-y)x²=4y⇒x²=4y/(1-y)≥0
y -∞ 0 1 +∞
4y --------------------------0++++++++++++++++++++
1-y+++++++++++++++++++++++0---------------------
4y(1-y) --------------------0+++++++|---------------------
y∈[0,1)=>Imf=[0,1)
Metoda 3
f'(x)=(8*x)/(x^2+4)^2
x -∞ 0 +∞
f(x) 1 desc 0 cresc 1
f'(x)---------------------0+++++++++++++ +++
limf(x) cand x->-infinit/+infinit=1
Imf=[0,1)
