âµf(x) = (2^x-a)/(2^x+1)æ¯å®ä¹å¨rä¸çå¥å½æ°
â´f(0)=(2^0-a)/(2^0+1)=0
â´a=1
f(x) = (2^x-1)/(2^x+1) = (2x+1-2)/(2^x+1) = 1 - 2/(2^x+1)
âµ2^x â
â´2^x+1 â
â´2/(2^x+1) â
â´1 - 2/(2^x+1) â
â´f(x)å¨Rä¸åè°å¢
å©ç¨å®ä¹è¯æå¦ä¸ï¼
令x1ï¼x2ï¼
f(x2)-f(x1) = { - 2/(2^x2+1) }- { - 2/(2^x1+1) }
= 2/(2^x1+1) - 2/(2^x2+1)
= 2{2^x2+1-2^x1-1}/{(2^x1+1)(2^x2+1)}
= 2{2^x2-2^x1}/{(2^x1+1)(2^x2+1)}
= 2x1{2^(x2-x1)-1}/{(2^x1+1)(2^x2+1)}
âµx2ï¼x1
â´x2-x1ï¼0
â´2^(x2-x1)ï¼1
â´2^(x2-x1)-1ï¼0
åï¼2x1ï¼0ï¼å¹¶ä¸{(2^x1+1)(2^x2+1)}ï¼0
â´2x1{2^(x2-x1)-1}/{(2^x1+1)(2^x2+1)}ï¼0
â´f(x2)ï¼f(x1)ï¼å¾è¯ã
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