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..

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)>=0y^2-8y+8>=0(y-4)^2>=8y=4+2根2 (1)令f(x)=x^2+(2-y)x+y-1且f(-1)f(1/2)...