Question

y=(x+3)/(x^2+5x+10)求值域

Answer 1

y=(x+3)/(x²+5x+10)

=(x+3)/[(x+3)²-(x+3)+4]

当x+3=0时 x=-3

当x≠-3时,

分子分母同时除以x+3得

=1/[(x+3)-1+4/(x+3)]

因为

(x+3)+4/(x+3)≥2√[(x+3)×4/(x+3)]=4

或

(x+3)+4/(x+3)≤-2√[(x+3)×4/(x+3)]=-4

所以,值域为 [-1/5,1/3]