定义在(-1,1)内的函数f(x)满足2f(x)-f(-x)=lg(x+1),求函数f(x)的解析式
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2f(x) -f(-x) = lg(x+1) (1)
put y = -x
2f(-y) - f(y) = lg(-y+1)
=> f(-y) = [lg(-y+1)+ f(y)] /2 (2)
from (1) and (2)
2f(x) - [lg(-x+1)+ f(x)]/2 = lg(x+1)
3f(x) = lg(x+1)+ lg(-x+1)
f(x) = (1/3) lg(x+1)(-x+1)
put y = -x
2f(-y) - f(y) = lg(-y+1)
=> f(-y) = [lg(-y+1)+ f(y)] /2 (2)
from (1) and (2)
2f(x) - [lg(-x+1)+ f(x)]/2 = lg(x+1)
3f(x) = lg(x+1)+ lg(-x+1)
f(x) = (1/3) lg(x+1)(-x+1)
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