F.4 函數...THX

The answer is as follows:

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2/(3+√2) + 2/( 3 - √2)
=[2(3-√2)+2(3+√2)]/[(3-√2)(3+√2)]
=(6-2√2+6+2√2)/(3-2)
=12/1
=12//

3.已知 f(x) = ax+ b, f(1) = -6 及 f(2) = 1

f(1) = -6
a+b=-6--(1)

f(2) = 1
2a+b=1--(2)

(2)-(1):
2a+b-a-b=1-6
a=-5
then b=-1//

so a=-5 , b=-1.

f( -5 )=a(-5)+b=(-5)(-5)-1=24//

4. 已知 f(x) = kx +2及 f(2) = 8.. 求值.
解方程 f [ f(x)] = 0

f(2) = 8
2k+2=8
k+1=4
k=3

f [ f(x)]=0
k[f(x)]+2=0
k(kx+2)+2=0
k^2 x +2k+2=0
9x+6+2=0
x=-8/9//


2008-10-19 21:11:10 補充：
5. 已知 F(x) = 1÷x+1及 g(x) =1÷x
a.求 f[g(x)]
b.解方程 f[g(x)] – f(x) = 5

f[g(x)]=1/[(1/x)+1]=x/(x+1)

f[g(x)] – f(x) = 5
x/(x+1)-1/(x+1)=5
(x-1)=5(x+1)
x-1=5x+5
4x=4
x=1//

2008-10-19 21:11:26 補充：
f[g(x)] – f(x) = 5
x/(x+1)-1/(x+1)=5
(x-1)=5(x+1)
x-1=5x+5
4x=-6
x=-3/2//