急急急拜託幫解國一函數問題

1.
(1)
總水量 = 50 × 30 × 1.8 立方公尺 = 2700 立方公尺
每小時抽水量 = (750/1000) × 60 = 45 立方公尺
經過 x 小時剩下的水量 = (2700 ‒ 45x) 立方公尺

f(x) = 2700 ‒ 45x

(2)
f(x) = (1/2)f(0)
2700 ‒ 45x = (1/2)(2700)
45x = 1350
x = 30

經過時間 = 30 小時


====
2.
當 x = 2 > ‒1 時： g(x) = x + 6
g(2) = 2 + 6 = 8

當 x = ‒3 < ‒1 時： g(x) = 5
g(‒3) = 5

g(2) ‒ g(‒3) = 8 ‒ 5 = 3


====
3.
(1)
f(0) = f(0 + 4) = f(4)
f(4) = f(4 + 4) = f(8)
f(8) = f(8 + 4) = f(12)
f(12) = f(12 + 4) = f(16)

所以f(0) = f(16)
f(0) ‒ f(16) = 0

(2)
f(3) = ‒7

f(3) = f(‒1+4) = f(‒1)
所以 f(‒1) = ‒7

f(‒1) = f(‒5 + 4) = f(‒5)
所以 f(‒5) = ‒7


====
4.
f(1) = 5
f(2) = f(1 + 1) = f(1) + 2 = 5 + 2 = 7
f(3) = f(2 + 1) = f(2) + 2 = 7 + 2 = 9
f(4) = f(3 + 1) = f(3) + 2 = 9 + 2 = 11

f(4) ‒ f(1) = 11 ‒ 5 = 6


====
5.
f(3) = (2 × 3 ‒3) ‒ 5 = ‒2
g(3) = (3/3) ‒ 3 = ‒2

f(3) + g(3) = (‒2) + (‒2) = ‒4


====
6.
f(x) + g(x) = 4x + 8 ...... [1]
f(x) ‒ g(x) = 2x ‒ 2 ...... [2]

[1] + [2]：
2f(x) = 6x + 6
f(x) = 3x + 3

[1] ‒ [2]：
2g(x) = 2x + 10
g(x) = x + 5

f(3) ‒ g(4)
= (3 × 3 + 3) ‒ (4 + 5)
= 12 ‒ 9
= 3


====
7.
Put x = 5

g(4)
= g(5 ‒ 1)
= f(5 + 7)
= f(12)
= 2 × 12 + 1
= 25


====
8.
f(9)
= f(2 × 3 + 3)
= f(3) + 7
= 6 + 7
= 13

f(21)
= f(2 × 9 + 3)
= f(9) + 7
= 13 + 7
= 20

f(45)
= f(2 × 21 + 3)
= f(21) + 7
= 20 + 7
= 27

奇摩知識沒限定說不能發問課業上的問題吧?

土扁大大花了不了時間，幫我解決了問題，是位好心人士。

土扁大大助長這種問作業的怪風氣，不會只是為了點數吧？！...