Question

求函数y=x2-x+2/x+1的值域

Answer 1

y=(x^2-x+2)/(x+1)

y=[(x+1)^2-3(x+1)+4]/(x+1)

y=(x+1)-3+4/(x+1)

令t=x+1 t≠0

f(t)=t+4/t-3

t>0时

f(t)≥2√t*4/t-3=4-3=1

t