中四數學函數(Functions)問題 急急跪求解答

1a.Put x = 1,
f(2) = 1 + 4 + k = 8
k = 3

1b.Put x = -2,
f(-1) = (-2)^2 + 4(-2) + 3
f(-1) = -1

1c.Replace x by x-1,
f(x) = (x-1)^2 + 4(x-1) + 3
= x^2 + 2x
Put f(x) = 3,
x^2 + 2x = 3
x^2 + 2x - 3 = 0
(x + 3)(x - 1) = 0
x = -3,1

2a.Put x = 1,
g(2) = 1 + 1/1^2
g(2) = 2

2b. Let y = x + 1/x,then
y^2 = x^2 + 2 + 1/x^2
x^2 + 1/x^2 = y^2 - 2
So g(y) = y^2 - 2
Replace y by x, g(x) = x^2 - 2
Put g(x) = -1,
x^2 - 2 = -1
x^2 = 1
x = +/-1

function f(x) such that f(x+1) = x^2+4x+k and f(2)=8.

a) f(x+1) = x^2+4x+k
f(2)=1^2 +4+k=8
5+k=8
k=3

b)f(x+1)=x^2+4x+3
f(-1)=0^2+4(0)+3=3

c)f(x)=(x-1)^2+4(x-1)+3
=x^2-2x+1+4x-4+3
=x^2+2x
hence
f(x)=3.
x^2+2x-3=0
(x+3)(x-1)=0
x=-3 or 1

2) It is given that g(x+1/x) = x^2 + 1/(x^2).
請問係o個FUNCTION OF G係(X+1)/X 定係X+ (1/X)???
a)Find the value of g(2).
x+1
b)Find the values of x such that g(x) = -1

2010-10-23 11:47:24 補充：
case 1 FUNCTION OF G係(X+1)/X
(X+1)/X =2
x+1 =2x
x=1
.'. g(2) = 1^2 +1/1=2

b) (X+1)/X =x
x^2=x+1

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


x+1/x=x

2010-10-23 11:57:38 補充：
1b)f(x+1)=x^2+4x+3
f(-1)=(-2)^2+4(-2)+3=-1


It is given that g(x+1/x) = x^2 + 1/(x^2).
a)Find the value of g(2).
x+1/x=2
x^2+1-2x=0
x=1(repeated root)

.'.g(2)= 1+1=2

2010-10-23 11:57:44 補充：
b)Find the values of x such that g(x) = -1
Let y = x + 1/x
y^2 = (x+1/x)^2
=x^2+2+1/x^2
x^2 + 1/x^2 = y^2 - 2
So g(y) = y^2 - 2
'.' g(x) = x^2 - 2
Put g(x) = -1,
x^2 - 2 = -1
x^2 = 1
x = +/-1