Given f [f (x)] = 4x - 1, find the analytic expression of the first order function f (x)

Given f [f (x)] = 4x - 1, find the analytic expression of the first order function f (x)

Let f (x) = ax + B, f [f (x)] = f (AX + b) = a (AX + b) + B = square a, x + AB + B = 4x-1, so square a = 4, a = ± 2, when a = 2, AB + B = 2B + B = - 1, B = - 1 / 3, f (x) = 2x-1 / 3, when a = - 2, AB + B = - 2b + B = - 1, B = 1, f (x) = - 2x + 1, f (x) can be expressed as (x) = 2x-1 / 3 or (x) = - 2x + 1

The mathematical function FX = x2-ax + A + 3, where x belongs to the closed interval - 2, 2 is used to find the maximum value of X

If - A / 2 ＜ - 2, then f (- 2) f (x) f (2) if - 2 ＜ - A / 2 ＞ 0, then f (- A / 2) f (x) f (2) if 0 ＜ - A / 2 ＜ 2, then f (- a / 2) f (x) f (- 2) if - A / 2 ＞ 2, then f (2) f (x) f (2) pure handwriting Mobile party

The quadratic function FX satisfies FX + 1-fx = 2x, and F0 = 1.1. Find the analytic expression of FX. 2. If GX = MX + 2, FX = FX GX. Find FX on [- 1,2]
The quadratic function FX satisfies FX + 1-fx = 2x and F0 = 1
1. Find the analytic expression of FX
2. If GX = MX + 2, FX = FX GX. Find the minimum value HM of FX on [- 1,2]
3. Find the minimum value of HM in M ∈ [- 1,2]

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From (1) we can get f (x) = x & # 178; - x + 1, from the question we can get FX = x & # 178; - x + 1-mx-2 = x & # 178; - (1 + m) X-1, function FX over fixed point (0, - 1). When 1 + m / 2 & lt; = - 1, take the minimum value f (- 1). When - 1 & lt; 1 + m / 2 & lt; 2, take the minimum value f (1 + m / 2). When 2 & lt; = 1 + m / 2, take the minimum value f (2)
&When (2) m ∈ [- 1,2], HM = f (1 + m / 2), let t = 1 + m / 2, M + 1 = 2t-1, so t ∈ [1 / 2,2], HM = f (T) = T & # 178; - (2t-1) T-1 = - T & # 178; = T-1, the following is the same as the second step