F (x) = SiNx, f [& x] = 1-x ^, find & X and its definition field. It's OK to solve ah, inverse function or other methods

F (x) = SiNx, f [& x] = 1-x ^, find & X and its definition field. It's OK to solve ah, inverse function or other methods

Because f (x) = SiNx, f [& (x)] = 1-x & # 178;
So because - 1 ≤ SiNx ≤ 1, so - 1 ≤ 1-x & # 178; ≤ 1,0 ≤ X & # 178; ≤ 2, - √ 2 ≤ X & # 178; ≤ √ 2
That is to say, the domain of definition of (x) is [- √ 2, √ 2]
And f [& x] = sin [& x] = 1-x & # 178;
So & (x) = arcsin (1-x & # 178;) + 2K π or & (x) = π - arcsin (1-x & # 178;) + 2K π, K ∈ Z

The domain of the inverse function f (x) = 3x (0 < x ≤ 2) is______ ．

∵ 0 ＜ x ≤ 2, ∵ 1 ＜ 3x ≤ 32 = 9, ∵ the definition field of inverse function of function f (x) = 3x (0 ＜ x ≤ 2) is (1,9]; so the answer is (1,9]

How to prove that the increase and decrease of function and inverse function are the same
Let y = f (x) be an increasing function, and prove that its inverse function F-1 (x) is also an increasing function

Because y = f (x) is an increasing function in its domain of definition, then y increases with the increase of X, that is, when X1 > X2, f (x1) > F (x2) and because there is an inverse function, then y corresponds to x one by one, then X1 > x2 can be obtained from F (x1) > F (x2), that is, for y = f ^ - 1 (x), if X1 > X2, f ^ - 1 (x1) > f ^ - 1 (x2)