函数y=x^2+ax-2/x^2-x+1的值域y≤2,求a的值.已知函数f(x)的定义域为R对任意XY都有f(x+y)=f(x)f(y),且当X＞0时,f(x)＞1. 证明 f(x)在R上递增
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函数y=x^2+ax-2/x^2-x+1的值域y≤2,求a的值.
y=(x^2+ax-2)/(x^2-x+1)
=[(x^2-x+1)+(a+1)x-3]/(x^2-x+1)
=1+[(a+1)x-3]/(x^2-x+1)
值域为(-∝,2],则
1+[(a+1)x-3]/(x^2-x+1)=1
f(x2)-f(x1)=f(x2)-f(x2+(x1-x2))=f(x2)-f(x2)*f(x1-x2)=f(x2)[1-f(x1-x2)]