Question

求函数y=(x²-2x-3)/(2x²+2x+1)的值域

Answer 1

函数y=(x²-2x-3)/(2x²+2x+1)的定义域为R

y=(x²-2x-3)/(2x²+2x+1)

＝>(2x²+2x+1)y=x²-2x-3

=>(2y-1)x²+2(y+1)x+y+3=0

=>△=b²-4ac=[2(y+1)]²-4(2y-1)(y+3)≥0

即-y²-3y+4≥0

所以（y+4）(y-1)≤0

=>-4≤y≤1

=>函数y=(x²-2x-3)/(2x²+2x+1)的值域[-4,1]