#ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#求f(x)=|x-1|+|x-2|+|x-3|+|x-4|+|x-15|的最小值?#ed_op#BR#ed_cl#若用大學生解法#ed_op#BR#ed_cl#令 g(x)=(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-15)^2#ed_op#BR#ed_cl#g'(x)=0,得x=5#ed_op#BR#ed_cl#5介於4和15之間,將4和15代入f(x)#ed_op#BR#ed_cl#f(4)=17#ed_op#BR#ed_cl#f(15)=50#ed_op#BR#ed_cl#所以x=4時,f(x)有最小值17#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#但x=3時,f(3)=16顯然比17小,17不是最小值#ed_op#BR#ed_cl#請問,哪裡出錯了?#ed_op#/DIV#ed_cl#大嘴 寫到:#ed_op#BR#ed_cl#G(x) = |x-1| + |x-2| + |x-9| 的最小值#ed_op#BR#ed_cl#令f(x)= (x-1)^2+(x-2)^2+(x-9)^2#ed_op#BR#ed_cl#微分f(x)= 2 ( (x-1)+(x-2)+(x-9) )令=0#ed_op#BR#ed_cl#3x-12 =0#ed_op#BR#ed_cl#x=4#ed_op#BR#ed_cl#x介於2與9間 將2與9代入G#ed_op#BR#ed_cl#G(2) =8#ed_op#BR#ed_cl#G(9) =15#ed_op#BR#ed_cl#得x在2值為8
#ed_op#/P#ed_cl##ed_op#P#ed_cl#f(x)= (x-1)^2+2(x-2)^2+3(x-3)^2+…+9(x-9)^2 #ed_op#/P#ed_cl##ed_op#P#ed_cl#為什麼不是當x=285/45=19/3時,f(x)有最小值f(19/3),非得取x為整數不可#ed_op#/P#ed_cl##ed_op#P#ed_cl#又若要求 f(x) = |x-1| + |x-2| + |x-9| 的最小值,若用您的「大學生解法」作的話,#ed_op#/P#ed_cl##ed_op#P#ed_cl#x為何值時?f(x)才有最小值?#ed_op#/P#ed_cl#大嘴 寫到:有兩種解法:#ed_op#BR#ed_cl#1.大學生解法#ed_op#BR#ed_cl##ed_op#BR#ed_cl#將題目改為: 求f(x)= (x-1)^2+2(x-2)^2+3(x-3)^2+…+9(x-9)^2 最小值#ed_op#BR#ed_cl##ed_op#BR#ed_cl#微分f(x)得 2 ( (x-1)+2(x-2)+3(x-3)+…+9(x-9) )令=0#ed_op#BR#ed_cl#(1+2+3---+9) x-(1+4+9---+81) =0#ed_op#BR#ed_cl#45 x=285#ed_op#BR#ed_cl#x=6.333333#ed_op#BR#ed_cl#當xε(6,7)時, 有最小值. x=6, 值85. x=7, 值82.#ed_op#BR#ed_cl##ed_op#/P#ed_cl##ed_op#P#ed_cl#