Question

y=x方/(x+2) [x>-2] 求值域

Answer 1

y=x^2/(x+2)

=(x^2-4+4)/(x+2)

=x-2+[4/(x+2)]

=x+2+[4/(x+2)]-4

≥2√(x+2)*[4/(x+2)-4

=0

当且仅当x=0时取等号

y=x方/(x+2) [x>-2] 求值域[0,+无穷)