g(x) = 3x+2, f(g(x)) = 7x-5 請問

Let z = g(x)

f(x) = 3x + 2
z = 3x + 2
3x = z - 2
x = (z - 2)/3

f(g(x)) = 7x - 5
f(z) = [7(z - 2)/3] - 5
f(z) = (7/3)z - (14/3) - 5
f(z) = (7/3)z - (29/3)

Since f(z) = (7/3)z - (29/3),
then f(x) = (7/3)x - (29/3)
(7/3)x = f(x) + (29/3)
x = (3/7) * [f(x) + (29/3)]
Hence, f⁻¹(x) = (3/7) * [x + (29/3)]

f⁻¹(2) = (3/7) * [2 + (29/3)]
f⁻¹(2) = (3/7) * (35/3)
f⁻¹(2) = 5

知識長 超厲害 給個讚

g(x)=3x+2，f(g(x))=7x－5請問，f^(－1)(2) 如何求得=5?
Sol
f(3x+2)=7x-5
f^(－1)(7x－5)=3x+2
Set 7x－5=2
x=1
f^(－1)(2)=3+2=5

g(x) = 3x+2, f(3x+2) = 7x-5

令3x+2=k , x=(k-2)/3.

f((k-2)/3)=7(k-2)/3-5=(7k-2)/3-9 => f(x)=(7x-2)/3 - 9

令(7x-2)/3 - 9 = y , [(y+9)3+2]/7=x => f(x)的反函數 = (3y+29)/7 = (3x+29)/7



(3x+29)/7 = 5 , 故x=2 . 合理.

希望對您有幫助.

2013-07-24 22:59:19 補充：
棒棒　我也喜歡知識長的做法.