標題:
lg既問題?
發問:
Suppose that y=lg(x/6) * lg(x/3), try to get the min. value of y? and figure out of x when the min. value of y occured. 更新: 雞尾包,唔該晒. but you answer is wrong.
最佳解答:
y=lg(x/6)*lg(x/3) =(lgx-lg6)(lgx-lg3) =(lgx)^2-(lg6+lg3)(lgx)+(lg6)(lg3) =(lgx)^2-(lg6+lg3)(lgx)+(lg6)(lg3) =(lgx)^2-(lg6+lg3)(lgx)+[(lg6+lg3)/2]^2-[(lg6+lg3)/2]^2+(lg6)(lg3) =[lgx-(lg6+lg3)/2]^2+(lg6)(lg3)-[(lg6+lg3)/2]^2 ∴when lgx=(lg6+lg3)/2, y reachs its min. value (lg6)(lg3)-[(lg6+lg3)/2]^2 ∴when y reachs its min. value, x=(6*3)^(1/2)=3√2
其他解答:
此文章來自奇摩知識+如有不便請留言告知
y = lg(x/6) * lg(x/3) = (lgx - lg6) (lgx - lg3) = (lgx)^2 - (lg6 + lg3) lgx + (lg6)(lg3) (lgx)^2 - (lg6 + lg3) lgx + (lg6)(lg3) = 0 has roots 6 and 3 min. value of y occured when x = (3 + 6)/2 = 9/2 y = lg(4.5/6) * lg(4.5/3) 自己計埋佢 ...
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