2 2x?2xlnxx?2x 000 00 所以g?x??g?x????x.(∵2lnx?x?4) 000 0 min x?2x?2 00 即m?x,所以fm?fx?2?2lnx?x?2?6,7,即6?fm?7. ???????? 0000 2xxx 3.已知函数fx?e?e?xe(e?2.718?,e为自然对数的底数),证明:fx存在唯一极大值点x, ???? 0 ln211 且??fx?. ?? 0 2 2e4 4e 答案:见解析 2xxxxx ? 解析:fx?e?e?xe,fx?e2e?x?2. ???? ?? xx 令hx?2e?x?2,则h?x?2e?1, ???? ∵时,?,在上为减函数; x????,?ln2?h?x??0h?x????,?ln2? ? x???ln2,???时,h?x??0,h?x?在??ln2,???上为增函数, 由于h?1?0,h?2?0,所以在?2,?1上存在x?x满足hx?0, ???????? 00 ∵hx在??,?ln2上为减函数, ???? ∴x???,x时,hx?0,即f?x?0,fx在??,x上为增函数, ?????????? 00 时,,即?,在上为减函数, x??x,0?h?x??0f?x??0f?x??x,0? 00 ? x??0,???时,h?x??0,即f?x??0,f?x?在?0,???上为增函数, 综上可知,fx存在唯一的极大值点x,且x??2,?1. ???? 00 x 0 ∵hx?0,∴2e?x?2?0, ?? 00 2 2 ?x?2??x?2?x?2x 2xxx0000 000 所以fx?e?e?xe??x?1??,, ????x???2,?1? 0000 ???? 224 ???? 2 x?2x11 ∵时,,∴; x???2,?1???f?x?? 0 444 1ln21 1?? ∵ln??2,?1,∴fx?fln??; ???? 0?? 2 2e2e2e 4e ?? ln211 综上知:??fx?. ?? 0 2 2e4 4e |
|