A. maths

First set x²-4x+3=0
(x-1)(x-3)=0
∴f(x)=0 when x=1 or 3, which is in the range of [0,5]
∴Min value of f(x)=0 as | x^2 - 4x +3 | must be greater than or =0
Consider tha shape of x²-4x+3, it is opening upward from the two sides, so it is obvious that the max. value occurs at the end pt.
f(0)=3
f(5)=8
∴Max. value of f(x)=8
∴range of f(x)=0 to 8

i can tell you how to find the corresponding range of f(x)

consider
g(x)=x^2 - 4x +3
=(x-3)(x-1)

when x=1 or 3
g(x)=0

Since the cofficient of the x^2 is 1 , the graph is upward

This means that g(0) or g(5) should be the largest one

Since g(5)=8>g(0)=3

And f(x)=|g(x)| should be >=0

so the corresponding range of f(x) is

0<=x<8

2008-06-07 21:20:41 補充：
&gt >
&lt

2008-06-07 21:21:28 補充：
YAHOO時常想人手機認證

2008-06-07 21:46:28 補充：
because x=1 and x=3 is belonged to the domain 0<5

So the minmimum value of f(x) is 0

2008-06-07 21:48:00 補充：
ＡＳ YOU KNOW THAT


第一個人當然無表達得０甘好﹐因為無參考資料。