A.Math 一問!!!!

5 -（　1 ／ x2 – 2x +2 )
let R=（　1 ／ x2 – 2x +2 )
「Min value=5 - max of R
in order to obtain max value or R, 分母要min
∴find the min value of x2 – 2x +2 By completing square」
consider y=x2 – 2x +2
= (x2 – 2x +12)+2-1 <-----12是1的2次方
=(x-1)2+1 <----(x-1)2是(x-1)的2次方
for x=1, min value of y=1
∴Min value=5-1/1
=4

5 -（　1 ／ x2 – 2x +2 )
let R=（　1 ／ x2 – 2x +2 )
「Min value=5 - max of R
in order to obtain max value or R, 分母要min
∴find the min value of x2 – 2x +2 By completing square」
consider y=x2 – 2x +2
= (x2 – 2x +12)+2-1 <-----12是1的2次方
=(x-1)2+1 <----(x-1)2是(x-1)的2次方
for x=1, min value of y=1
∴Min value=5-1/1
=4

f(x) = 5 - 1 / (x^2 + 2x + 2)
df(x)/dx = -(-1)[(x^2 + 2x + 2)^(-2)] (2x + 2)

local maximum or minimum appear when df(x)/dx = 0

0 = (2x + 2) / (x^2 + 2x + 2)^2
0 = 2x + 2
0 = x + 1
x = -1

2007-01-02 19:48:07 補充：
睇錯左正負方法都係d佢一次set f `(x) = 00 = (2x - 2) / (x^2 - 2x + 2)^2x = 1when x = 0 & x = 2f `(0) = -1/2 >> -vef `(2) = 1/2 >> +vef(1) is a min ptf(1) = 4

首先5可以不理，因為constant不會隨x而變

x^2-2x+2=x^2-2x+1+1
= (x-1)^2+1

So x^2-2x+2 最小是1
1/( x^2-2x+2) 最大是1 (想想便知)

5-1/( x^2-2x+2) 最小是4, 當且僅當x=1

2007-01-02 19:20:26 補充：
Remarks: this is only true for real xAlso, we can prove it by inequalitiesx^2-2x 2=(x^2-2x 1) 1=(x-1)^2 1 = 1 -1/(x^2-2x 2)