Question

f(x)=[6-5x]/(1-x)^2 的值域

Answer 1

y=[6-5x]/(1-x)²

y(1-x)²=6-5x

yx²-2yx+y=6-5x

yx²+(5-2y)x+y-6=0,

△=(5-2y)²-4y(y-6)=25-20y+24y=4y+25≥0,

即y≥-25/4,

即f(x)=[6-5x]/(1-x)^2 的值域为[-25/4,+∞).