急 !急 ! 急 ! F4 Function 計法

1a.
given f(1+x) = f(1)+f(x)
when x = 0, f(1+0) = f(1) + f(0)
=> f(0) = 0
1b.
when x=-1, f(1+(-1)) = f(1) + f(-1)
f(0) = f(1) + f(-1)
at 1a. f(0) = 0, thus f(-1) = - f(1)

2.
2f(x) = 2(2x-1) = 4x - 2
f(x+1) = 2(x+1) - 1 = 2x + 2
2f(x) = f(x+1)
4x-2 = 2x + 2
2x = 4
x = 2

10a.
quantity = 3000 - 100x
profit per toy = x - 20
total profit, P = (x-20)(3000-100x)
10b.
P = (x-20)(3000-100x) = -100x^2 + 5000x + 60000
d(P)/dx = P' = -200x + 5000
d(P')/dx = P'' = -200 < 0 => P will provide a maximum value when P' = 0
=> x = 25
10c.
when x = 25, quantity = 3000 - 100(25) = 500

11a.
q = 12000 - 50(p-160)
11b.
Income, I = p q = p [12000 -50(p-160)]
I = 20000p - 50p^2
d(I)/dp = I' = 20000 - 100p
d(I')/dp = I'' = -100 < 0 => I will provide a maximum value when I' = 0
=> p = 200
thus, I = 2,000,000

p.s. for a quadratic equation y = Ax^2 + Bx + C, the curve opens downwards when A < 0, such that a maximum value of y will get at x = -B/2A, and vice versa.


2011-11-15 17:44:09 補充：
Q. 請問第2題點解是f(x+1) = 2(x+1) - 1
而不是f(x+1) = 2x+1 - 1

A. f(x) = 2x - 1
=> f(z) = 2z - 1
put z = x+1 => f(x+1) = 2(x+1) - 1......ok??

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