Let f (x) = 3x, f (a) = 2, G (x) = 3ax-4x. (1) find the analytic expression of G (x); (2) find the range of G (x) when x ∈ [- 2, 1]

Let f (x) = 3x, f (a) = 2, G (x) = 3ax-4x. (1) find the analytic expression of G (x); (2) find the range of G (x) when x ∈ [- 2, 1]

(1) Let's get 3A = 2 from F (a) = 2, a = log32, a = 2, a = log32, and let's get 3A = 2, a = log32, and the (3) g (x) = (3 log32) x (3a) x-4x = (3 log32) x {(3 log32) x-4x = (3 log32) x-4x (3 log32) x-4x = (3 log32) x-4x-4x = (3 log32) x {(3 log32) x-4x = 4x = 4x = 2x-4x-4x-4x-4x) 2x (2x) 2 (2) let 2x = 2x = 2x (2) 2 = 2x) 2, and (2x) 2) 2 = 2, and let let's be the domain of the domain of \\\\\\\\]

Given the function f (x) = 2x, f (a + 2) = 12, G (x) = 2ax-9x, the domain of definition of G (x) is [0,1]. (I) find the analytic expression of G (x); (II) find the range of value of G (x)

(I) let t = 3x, X ∈ [0, 1], then t ∈ [1, 3] ｜ g (T) = t-t2 = - (t − 12) 2 + 14. According to the properties of quadratic function, G (T) = t-t2 = - (t − 12) 2 + 14 decreases monotonically on [1, 3]. When t = 1, G (1) = 0, when t = 3, G (3) = - 6 ｜ the range of function is [- 6, 0]

Given the function f (x) = x ^ 2, X ∈ [- 2,2] and the function g (x) = AX-1, X ∈ [- 2,2], if for any X1 = ∈ [- 2,2], there always exists x0 ∈ [- 2,2]
Let g (x0) = f (x1) hold, then the value range of real number a is

From F (x) = x ^ 2, X ∈ [- 2,2], two endpoints a (- 2,4) B (2,4) of the function are obtained
The function g (x) = AX-1, X ∈ [- 2,2], if for any X1 = ∈ [- 2,2], there is always x0 ∈ [- 2,2] such that G (x0) = f (x1) holds
It shows that G (x) and function f (x) = x ^ 2, X ∈ [- 2,2] must have intersection
Since g (x) always passes a certain point C (0, - 1)
The slope of CB is (4 + 1) / (2-0) = 5 / 2
The slope of Ca is (4 + 1) / (- 2-0) = - 5 / 2
The slope of the function g (x) = AX-1 is a
-5/2