Question

求函数y=(x^2+2x-1)/(x-1),x∈[-1,1/2]的值域,用二次方程根的方法..

Answer 1

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

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

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

x^2+(2-y)x+y-1=0

关于x的方程有实数根,判别式大于等于零

（2-y)^2-4(y-1)>=0

y^2-8y+8>=0

(y-4)^2>=8

y=4+2根2 (1)

令f(x)=x^2+(2-y)x+y-1

且f(-1)f(1/2)