Let x be a real number. Find the minimum value of root(x^2-4x+13)+root(x^2-14x+130).?
回答 (2)
2016-03-20 12:18 am
✔ 最佳答案
Solx^2－4x+13=(x－2)^2+3^2
x^2－14x+130=(x－7)^2+9^2
題目改為
求兩點(2，3)，(7，9)過x軸最短距離
(2，3)在x軸對稱點(2，－3)
(7－2)^2+(9+3)^2
=25+144
=169
最小值=13
2016-03-20 3:51 pm
y=root(x^2-4x+13)+root(x^2-14x+130)
dy/dx=0.5(2x-4)(x^2-4x+13)^(-0.5) +0.5(2x-14)(x^2-14x+130)^(-0.5)
dy/dx=(x-2)(x^2-4x+13)^(-0.5) +(x-7)(x^2-14x+130)^(-0.5)
dy/dx|x=0 =-2(13)^(-0.5)-7(130)^(-0.5)
最小值=16.51780446
dy/dx=0.5(2x-4)(x^2-4x+13)^(-0.5) +0.5(2x-14)(x^2-14x+130)^(-0.5)
dy/dx=(x-2)(x^2-4x+13)^(-0.5) +(x-7)(x^2-14x+130)^(-0.5)
dy/dx|x=0 =-2(13)^(-0.5)-7(130)^(-0.5)
最小值=16.51780446
收錄日期: 2021-04-18 14:47:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160319064037AAthOmb