9.f(x)=(1/3-x)^1/2+(x-1/5)^1/2的最大值為a,最小值為b,則(a,b)=_______
1/5 <= x <= 1/3
f(x)=(1/3-x)^1/2+(x-1/5)^1/2 =>
[f(x)]^2=(1/3-x)+(x-1/5)+(-x^2+8*x/15-1/15)^1/2
=2/15+[-(x-4/15)^2+1]^1/2=>
[f(x)]^2 MAX x= 4/15 [f(x)]^2=2/15+1=17/15 a=(17/15)^1/2
[f(x)]^2 MIN x=19/15 [f(x)]^2=2/15+0=2/15
but 1/5<x<1/3 所以x=1/3 [f(x)]^2=[2+4*(14)^1/2]/15
b={[2+4*(14)^1/2]/15}^1/2
(a,b)=((17/15)^1/2,{[2+4*(14)^1/2]/15}^1/2)
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 60.245.65.194
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 111.251.27.72