求Y=x/x²+2x+1的值域

问题描述:

求Y=x/x²+2x+1的值域

答:
y=x/(x²+2x+1)
=(x+1-1)/(x+1)²
=1/(x+1)-1/(x+1)²
= - [1/(x+1) -1/2 ]² +1/4
=1/4
所以:
值域为(-∞,1/4]