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æ³ã令 â(x-1) = tï¼å x = t�0�5+1ï¼äºæ¯ y = t/(t�0�5+1). å 为 t�0�5+1â¥2tï¼æ以 t/(t�0�5+1) ⤠1/2ï¼åæ¾ç¶ t/(t�0�5+1) ⥠0ï¼æ以 0â¤t/(t�0�5+1)â¤1/2.å³ y = â(x-1) /x çå¼åæ¯ [0ï¼1/2].
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