Question

求函数的值域：y=(x^2+4x+5)/(x+1)

Answer 1

y=(x^2+4x+5)/(x+1)

y(x+1)=(x^2+4x+5)

x^2+(4-y)x+5-y=0

△=(4-y)^2-4(5-y)=y^2-4y-4≥0

y≥2+2√2,或,y≤2-2√2

值域：(-∞,2-2√2]U[2+2√2,+∞)