1.令y=x=0,å¾f(0)=f(0)+f(0)=2f(0),å¾f(0)=0
å令y=-x,å¾f(0)=f(x)+f(-x)=0,å¾f(-x)=-f(x)
æ以f(x)å¨Rä¸ä¸ºå¥å½æ°
2.设x1,x2为Rä¸ä»»æ两å®æ°ï¼ä¸x1<x2,åx2-x1>0,æ以f(x2-x1)<0
ç±f(x+y)=f(x)+f(y)ç¥
f(x2)=f(x1)+f(x2-x1)
æ以f(x2)-f(x1)=f(x2-x1)<0
å³f(x2)<f(x1)
æ以f(x)å¨Rä¸ä¸ºåå½æ°
æè§ç»åºçf(3)=-2没æç¨ä¸ï¼æ¯å¦è¿æå«çé®æ²¡ååºæ¥åï¼
f(x)å¨[-12,12]ä¸çæ大å¼ä¸ºf(-12)
æå°å¼ä¸ºf(12)
f(3)=-2.f(x+y)=f(x)+f(y),
f(12)=2*f(6)=4*(f3)=-8
f(-12)=-f(12)=8
æ以
f(x)å¨[-12,12]ä¸çæ大å¼ä¸º8
æå°å¼ä¸º-8
åæäºä¸ä¸ï¼è¿éæåèçæ¡ã
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