Question

用分离常数法解函数值域求讲解!1.y＝x＋3／x＋12.y＝x/x+1

Answer 1

1. y=(x+3)/(x+1)

=(x+1+2)/(x+1)

=1+2/(x+1)

因为2/(x+1)的值域不为0,

因此y的值域不为1

2. y=x/(x+1)

=(x+1-1)/(x+1)

=1-1/(x+1)

因为1/(x+1)的值域不为0

故y的值域不为1.