#ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(x)=2x^2+(a+1)x-a(a-1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#when a=2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(x)=2x^2+3x-2=(2x-1)(x+2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#-2<x<1/2 為負數#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#路人甲 寫到:#ed_op#DIV#ed_cl#這樣討論是不行地#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#必須配合圖形才才完整#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#若取a=2,明顯也成立#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(x)=2x^2+3x-6=2(x+4/3)^2-57/8 <----有問題#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(1)=-1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(0)=-6#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#若0<x<1,由圖形可知f(x)<0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#