f(x)=-t+1/2t+2的值域怎么求 函数f(t)=t+根号（1-2t）的值域是？答案是（负无穷大...
1.\u5df2\u77e5f(x)=x^2-2x+1(x\u2208[t,t+2])\uff0c\u5206\u522b\u6c42\u51fa\u503c\u57df\uff0c\u6700\u5927\u503c\uff0c\u6700\u5c0f\u503c\u3002 \u9898\u76ee\u672a\u5b8c\u5728\u4e0b\u9762f(x)=x²-2x+1=(x-1)²\uff0cx\u2208[t,t+2]\u5bf9\u79f0\u8f74\u4e3ax=1\uff0c\u5f53x|1-t-2|\uff0c\u5373-1<t<0\u65f6\uff0cf(x)max=f(t)=t²-2t+1\u3002
f(x)=x²+x+1-a=(x+1/2)²+3/4-a\uff0cx\u2265a\u5bf9\u79f0\u8f74\u4e3ax=-1/2\u5f53a\u2264-1/2\u65f6\uff0cf(x)min=f(-1/2)=3/4-a\uff1b\u5f53a>-1/2\u65f6\uff0cf(x)min=f(a)=a²+a+1-a=a²+1
\u4f7f\u7528\u6362\u5143\u6cd5
\u4ee4\u6839\u53f7\uff081-2t\uff09=x
\u5f97t=\uff081-x*\uff09/2
\u5f97\u4e00\u4e8c\u6b21\u51fd\u6570
\u5373\u5f97\u7b54\u6848
定义域为t≠0，故有一条垂直渐近线t=0
而t->+0时，有 f(t)->+∞
t->-0时，有 f(t)->-∞
又f'(t)=-1-1/(2t^2)<0
∴f(t)在定义域上为单调递减函数
A=lim(t->∞) [f(t)/t]
=lim(t->∞) [-1+1/(2t^2)+2/t]
=-1
B=lim(t->∞) [f(t)-At]
=lim(t->∞) [-t+1/(2t)+2+t]
=lim(t->∞) [1/(2t)+2]
=2
∴f(t)还有一条斜渐近线y=Ax+B=-x+2
∴f(t)的值域为(-∞,+∞)
解：函数在(-∞，0)和（0，+∞）上均为单调递减，故不存在最小值与最大值，
所以函数的值域为R
f(x)=(1-t)/(2t+2)=1/(1+t)-1/2
由该图象由反比例函数f(x)=1/t向下平移1/2再向左平移1得到，因而y不等于-1/2，值域为(-∞，-1/2)U（-1/2，+∞）
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