Question

f(x)=(x^2+2x+1)/(x-1)的值域

给的答案为 负无穷到0的左开右闭区间
并上 8到正无穷的左闭右开区间
求详细过程，各位学霸，摆脱了哈~

Answer 1

答：
f(x)=(x^2+2x+1)/(x-1)
=[(x-1)^2+4(x-1)+4]/(x-1)
=(x-1)+4/(x-1)+4…………利用对勾函数性质解答如下：
1）当x-1>0时：
f(x)=(x-1)+4/(x-1)+4>=2√4+4=8
2）当x-1<0时：
f(x)=(x-1)+4/(x-1)+4<=-2√4+4=0
所以：f(x)的值域为（-∞，0]∪[8,+∞）

Answer 2

f(x)=(x^2+2x+1)/(x-1)
= x+3 + 4/(x-1)
f'(x) = 1-4/(x-1)^2 =0
(x-1)^2 -4 =0
x^2-2x-3=0
(x-3)(x+1)=0
x=3 or -1

f''(x) = 8/(x-1)^3
f''(3)>0 (local min)
f''(-1) <0 (local max)

f(-1) = -1+3 +4/(-2) =0
f(3) = 3+3 + 4/2 = 8

值域 =(-∞, 0] or [8,+∞)

Answer 3

Answer 4

y=f(x)=(x^2+2x+1)/(x-1)
x^2+(2-y)x+y+1=0，方程x有解则
△=(2-y）^2-4(y+1)>=0
即y∈（-∞，0]∪[8,+∞）