關於函數 20點

6)已知f(x)=3x-2
b)驗證[f(2)]^2≠f(2^2)
[f(2)]^2 = [3*2 - 2]^2
= (6 - 2)^2
= 16
f(2^2) = 3(2^2) - 2
= 12 - 2
= 10
Since 16=/=10
Hence, [f(2)]^2≠f(2^2)
c)判斷f(√ 4)=√ f(4)是否成立
[f(√ 4)]^2 =/= f((√ 4)^2), from (b)
f(√ 4)=/=√ f(4)
7)已知G(x)=x(2x-1)
a)求G(2)和G(1/2)的值
G(2) = 2*(4-1) = 6
G(1/2) = (1 - 1)/2 = 0
b)判斷G(1/2)=1/G(2)是否成立
G(1/2) = 0
1/G(2) = 1/6
0=/=1/6
Hence, G(1/2)=/=1/G(2)

8)若f(x)=x(x+1),求下列各值
a)2f(-4)
2f(-4) = 2[(-4)(-4+1)]
= 2(4)(3)
= 24
b)f(2)+[f(-3)]^2
= 2(2+1) + [(-3)(-3+1)]^2
= 6 + (3^2)(2^2)
= 6 + 9*4
= 42

9)若f(t)=t^2+1及g(t)=2t,求下列各值
a)g(3)+f(0)
= 2*3 + 0^2 + 1
= 7
b)f(1)xg(-1)
= (1^2 + 1)(2*-1)
= -2(2)
= -4
c)g(4)/f(3)
= (2*4)/(3^2 + 1)
= 8/10
= 4/5