1.
求f(x)=5(x-1)^2+3(x-3)^2+2(x-4)^2+2(x-7)^2在x=a時有最小值b,求數對(a,b)
2.若-1<=x<=3,則f(x)=-x^2+3x+1的最大值為a最小值為b,求數對(a,b)=?
需要詳細解析
多項式函數
2011-10-09 5:36 pm
回答 (3)
2011-10-09 6:32 pm
✔ 最佳答案
1.f(x)=5(x-1)^2+3(x-3)^2+2(x-4)^2+2(x-7)^2==>f(x)=12x^2-72x+162=12(x^2-6x)+162=12(x-3)^2+54所以數對(a,b)=(3,54)
2.
f(x)=-x^2+3x+1=-(x^2-3x-1)=-(x-(3/2))^2+13/4
==>f(-1)=-3 ,f(0)=1,f(1)=3 ,f(2)=3,f(3/2)=13/4所以數對(a,b)=(13/4,-3)
2011-10-11 5:44 pm
這個解法很能理解,棒!
2011-10-09 6:44 pm
－x^2+3x+1
=－(x^2－3x)+1
=－(x^2－2*x*3/2+9/4)+1+9/4
=－(x－3/2)^2+13/4
－1<=x<=3
－1－3/2<=x－3/2<=3－3/2
－5/2<=x－3/2<=3/2
0<=(x－3/2)^2<=25/4
－25/4<=－(x－3/2)^2<=0
－25/4+13/4<=－(x－3/2)^2+13/4<=13/4
－3<=-x^2+3x+1<=13/4
(a，b)=(13/4，－3)
=－(x^2－3x)+1
=－(x^2－2*x*3/2+9/4)+1+9/4
=－(x－3/2)^2+13/4
－1<=x<=3
－1－3/2<=x－3/2<=3－3/2
－5/2<=x－3/2<=3/2
0<=(x－3/2)^2<=25/4
－25/4<=－(x－3/2)^2<=0
－25/4+13/4<=－(x－3/2)^2+13/4<=13/4
－3<=-x^2+3x+1<=13/4
(a，b)=(13/4，－3)
收錄日期: 2021-05-02 10:44:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111009000015KK01780