设fx为奇函数
\u8bbe\u51fd\u6570f'x\u662f\u5947\u51fd\u6570fx x\u2208R\u7684\u5bfc\u51fd\u6570\uff0cf(-1)=0,\u5f53x\uff1e0\u65f6\uff0cxf'x-fx\uff1c0\uff0c\u5219\u4f7f\u5f97ffx\u662f\u5947\u51fd\u6570,hx=f(x+ 2)\u53ef\u4ee5\u662f\u4e3a\u5947\u51fd\u6570\uff0c\u4ea6\u53ef\u4ee5\u4e0d\u4e3a\u5947\u51fd\u6570\uff0c
\u4f8b\u5982f(x)=tan\u03c0x\u662f\u5947\u51fd\u6570\uff0c\u800cf(x+2)\u4e5f\u662f\u5947\u51fd\u6570
f(x)=x\u662f\u5947\u51fd\u6570\uff0cf(x+2)=x+2\u4e0d\u662f\u5947\u51fd\u6570
所以f(-x)=-f(x)
所以当x小于0时，f(x)=-x(1+x)
解：令x<0,则，-x>0,代入f(x)=x(1+x),得
f(-x)=-x(1-x)
因为f(-x)=-f(x)
所以-x(1-x)=-f(x)
得，f(x)=x(1-x)，此时，x<0
选择B
所以当x小于0时,f(x)=-x(1-x)
A -x(1-x) B x(1-x) C -x(1+x) D x(1+x)
=A+f(x)Bx(1-x)C+f(x)Dx(1+x)
=A+f(x)[Bx(1-x)C+Dx(1+x)]
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