设fx=x/（x²+1),求f(1/x)解析式
f\uff08x\uff0b1\uff09\uff1dx²\uff0b4x\uff0b1\u6c42fx\u89e3\u6790\u5f0ff\uff08x\uff0b1\uff09\uff1dx²\uff0b4x\uff0b1=x²+4x+4-3=(x+2)²-3=[(x+1)+1]²-3
\u6240\u4ee5f(x)=(x+1)²-3=x²+2x+1-3=x²+2x-2
\u91c7\u7eb3\uff0c\u7ed9\u597d\u8bc4\uff0c\u70b9\u91c7\u7eb3
\u89e3\uff1a
\u5206\u5f0f\u6709\u610f\u4e49\uff0cx\u22600
f(x-1/x)=(x +1/x)²=(x-1/x)²+4
\u5c06x-1/x\u6362\u6210x
f(x)=x²+4
\u51fd\u6570\u89e3\u6790\u5f0f\u4e3af(x)=x²+4 (x\u22600)
x\u22600\uff0cx+1\u22600 x\u2260-1
f(x+1)=(x+1)²+4=x²+2x+5
\u51fd\u6570f(x+1)\u7684\u89e3\u6790\u5f0f\u4e3af(x+1)=x²+2x+5 (x\u2260-1)
\u4f60\u53e6\u5916\u5199\u7684\u540c\u4e00\u9053\u9898\uff0c\u5df2\u7ecf\u56de\u7b54\u8005\u7684\u9519\u8bef\u5728\u4e8e\u6ca1\u6709\u8003\u8651x\u7684\u53d6\u503c\u8303\u56f4\u3002
解：令1/x=u，则f(x)=x/(x²+1)=(1/x)/[1+(1/x)²]，故f(u)=u/(1+u²)，即f(1/x)=(1/x)/[1+(1/x)²].
f(x)=x/(x²+1)
f(1/x)=(1/x) / (1/x²+1)
分子：1/x
分母1/x²+1
分子分母同乘以x²得：
f(1/x)=x/(x²+1)
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