å·²ç¥f(x)=(ax+b)/(x²+1)为å®ä¹å为Rä¸çå¥å½æ°ï¼ä¸f(1)=1/2.
ï¼1ï¼æ±f(x)ç解æå¼ï¼ï¼2ï¼è¯æy=f(x)å¨ï¼-1,0ï¼ä¸çåè°æ§.
解ï¼(1)ãç±äºf(x)æ¯å¥å½æ°ï¼æ
å¿
æb=0ï¼åf(1)=a/2=1/2ï¼æ
a=1ï¼å³f(x)=x/(x²+1).
(2)ã设-1<x₁<x₂<0æ¯åºé´(-1ï¼0)ä¸çä»»æ两ç¹ã
ç±äºf(x₁)-f(x₂)=[x₁/(x₁²+1)]-[x₂/(x₂²+1)]=(x₁x₂²+x₁)-(x₂x₁²+x₂]/[(x₁²+1)(x₂²+1)]
=[x₁x₂(x₂-x₁)+(x₁-x₂)]/[(x₁²+1)(x₂²+1)]=[(x₁-x₂)(1-x₁x₂)]/(x₁²+1)(x₂²+1)]<0
è¿æ¶å 为-1<x₁<x₂<0ï¼æ
x₁-x₂<0ï¼0<x₁x₂<1ï¼1-x₁x₂>0ï¼x₁²+1>0ï¼x₂²+1>0ã
â´f(x₁)<f(x₂)ï¼å³f(x)å¨åºé´(-1ï¼0)å
åè°å¢ã
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