高二數學 急求解 拜託了?

f(x) = 2[1+sin(3x+π/6)]

(1)
∵ -1 ≦ sin(θ) ≦ 1,
∴ 0 ≦ 1 + sin(θ) ≦ 2
∴ 0 ≦ 2(1+sin(θ)) ≦ 4
∴ 0 ≦ f(x) ≦ 4

(2)
sin(θ) 在 π/2 時得最大值 1,
∴ sin(3．π/9 + π/6) 達最大值 1.
即 f(π/9) = 4 得最大.

(3)
sin(t) 週期 2π,
sin(t+a) 週期也是 2π.
sin(kt+a) 週期則是 2π/k.
∴ sin(3x+π/6) 週期是 2π/3
∴ f(x) 週期也是 2π/3,
f(x+2π/3) = 2[1+sin(3(x+2π/3)+π/6)]
    = 2[1+sin(3x+2π+π/6)]
    = 2[1+sin(3x+π/6)]
    = f(x) for all x in R.

(4)
sin(3x+π/6) 之振幅為 1,
故 f(x) = 2[1+sin(3x+π/6)] 之振幅為 2.

(1), (4) 正確.