a.maths ~~

取值範圍=the range of values of ...

(i)其實唔需要用到微分去做呢條數,咁做係將簡單複雜化
1.completing square
f(x)=x^2-6x+16
=x^2-6x+(6/2)^2-(6/2)^2+16
=x^2-6x+(3)^2+7
=(x-3)^2+7
the minimum value of f(x) is 7,
since f(x)has no real roots(delta<0),
the answer is f(x) > 7

(ii)g(x)=2002/f(x)
since f(x)is>7
g(x)>2002/7
g(x)>286

1.
df/dx = 2x - 6.
Put df/dx = 0, we get x = 3.
When x = 3 , y = 9 - 18 + 16 = 7.
Therefore, the range of f(x) is greater than 7 because 7 is the minimum.
2.
g = 2002/f
gf = 2002
g(x^2 -6x + 16) = 2002
gx^2 - 6gx + (16g - 2002) = 0
For x to be real, delta > 0
So 36g^2 - 4g(16g - 2002) > 0
36g^2 - 64g^2 + 8008 > 0
28g^2 - 8008 < 0
g^2 - 286 < 0
[g + sqrt286][g - sqrt286] < 0
Therefore, the range of g(x) is -sqrt286 < g(x) < sqrt286.

2008-08-11 17:53:30 補充：
Correction: The range of g(x) should equal or greater and equal or less than because delta is greater or equal to zero.

2008-08-11 17:57:33 補充：
Correction: Line 3 should be f(x) = 7, not y=7, sorry.

2008-08-11 20:13:49 補充：
Correction: For Q2, there is a mistake at line 7, it should be 36g^2 - 64g^2 + 8008g > 0, so the result should be g(g- 286) < 0, that is the range of g(x) is 0< g(x) < 286. A range of g< 286 is not 100% correct, g must be greater than zero also. Sorry for the mistake.