解ï¼å·²ç¥ä¸æ¬¡å½æ°Y=KX+B(Kä¸çäº0ï¼ç»è¿ï¼1ï¼2ï¼
ä¸å½X=-2æ¶ï¼Y=-1 ï¼
å°åæ ç¹ä»£äººä¸æ¬¡å½æ°Y=KX+Bå¾ï¼
2=k+b
-1=-2k+b
â´K=1ï¼b=1
ä¸æ¬¡å½æ°Y=KX+Bå°±çäºY=x+1.
P(A,Bï¼æ¯æ¤ç´çº¿ä¸å¨ç¬¬äºè±¡éå
çä¸ä¸ªå¨ç¹
ä¸PB=2PA;åPç¹çåæ å°±æ¯P(2PA ,PA),
å°Pç¹åæ 代人Y=x+1.å¾
PA=±1
PB=±2
å 为P(A,Bï¼æ¯æ¤ç´çº¿ä¸å¨ç¬¬äºè±¡éå
çä¸ä¸ªå¨ç¹åï¼
PA=1ï¼PB=-2
æ以Pç¹åæ æ¯Pï¼-2ï¼1ï¼
f(x)å®ä¹åx>-1ä¸xâ 0
f(x)=1/[(x+1)ln(x+1)]
f`(x)=-[(x+1)ln(x+1)]`/[(x+1)ln(x+1)]^2
=-[ln(x+1)+1]/[(x+1)ln(x+1)]^2
åæ¯[(x+1)ln(x+1)]^2>0
åªé讨论-[ln(x+1)+1]çæ£è´
å½-[ln(x+1)+1]â¥0æ¶
-1<xâ¤1/e-1
æ¤æ¶f`(x)â¥0
å½-[ln(x+1)+1]<0æ¶
x>1/e-1
æ¤æ¶f`(x)<0
â´f(x)çå¢åºé´(-1,1/e-1]
ååºé´[1/e-1,+â)
温馨提示:答案为网友推荐,仅供参考