If the domain of function f (x2 + 2) is [0,2], find the domain of y = f (x)

The domain of F (X & # 178; + 2) is [0,2], that is, 0 ≤ x ≤ 2
So 0 ≤ X & # 178; ≤ 4, so 2 ≤ X & # 178; + 2 ≤ 6
So the domain of F (x) is [2,6]
Note that the domain of a function must be the value range of an independent variable

The following four functions are: (1) y = x + 1; & nbsp; (2) y = X-1; & nbsp; (3) y = x2-1; & nbsp; (4) y = 1X, where ()
A. （1）（2）B. （1）（2）（3）C. 2）（3）D. （2）（1）（4）

(1) The definition and range of y = x + 1 are real number set R, so the definition and range are the same; (2) the definition and range of y = X-1 are real number set R, so the definition and range are the same; (3) the definition of y = x2-1 is real number set R, and the range is [- 1, + ∞), so the definition and range are not the same; (4) the function y = LX

If the definition field of function y = x2-3x-4 is [0, M], and the value range is [− 254, − 4], then the value range of M is ()
A. （0，4]B. [−254，−4]C. [32，3]D. [32，+∞)

Y = x2-3x-4 = x2-3x + 94-254 = (x-32) 2-254, the definition field is [0, M], then when x = 0, the function value is the largest, that is, y max = (0-32) 2-254 = 94-254 = - 4, and the value field is [- 254, - 4], that is, when x = m, the function is the smallest and Y min = - 254, that is, - 254 ≤ (m-32) 2-254 ≤ - 40 ≤ (m-32) 2 ≤ 94, that is, m ≥ 32 (1), that is, (m-32) 2 ≤ 94m-32 ≥ - 332 and m-32 ≤ 320 ≤ m ≤ 3 (2) So: 32 ≤ m ≤ 3, so choose C

Find the definition field of the function f (x) = (x-1) &# 186; + (2-x) ^ & # 189

The base of power 0 is not 0,
1 / 2 power is the arithmetic square, the base is non negative
Some inequalities are obtained
X-1≠0
2-X≥0,
The domain of definition is x ≤ 2 and X ≠ 1

The definition field of function y = (x + 1) & # 186; / √ (1-2 ^ x) is

X + 1 ≠ 0, and 1-2 ^ x ＞ 0
So: x < 0 and X ≠ - 1

Y = [f (x)] and # 186;, then --- (solution of function definition field)

Y = [f (x)] &186; meaningful
If f (x) ≠ 0 and f (x) itself is meaningful, it depends on the specific analytical formula of F (x)

The known function f (x) satisfies the following conditions: (1) for any x ∈ (0, + ∞), f (2x) = 2F (x) holds; (2) when x ∈ (1,2], f (x) = 2-x. if f (a) = f (2020), then the minimum positive real number a satisfying the condition is______ ．

Let x ∈ (2m, 2m + 1), then x2m ∈ (1, 2]; f (x2m) = 2-x2m, then f (x) = 2F (x2) = =2mf (x2m) = 2m + 1-x, where M = 0, 1, 2 F (2020) = 210f (20201024) = 211-2020 = 28 = f (a) let a ∈ (2m, 2m + 1), then f (a) = 2m + 1-A = 28  a = 2m + 1-28 ∈ (2m, 2m + 1), i.e. m ≥ 5, i.e. a ≥ 36. The minimum positive real number a satisfying the condition is 36, so the answer is: 36

It is known that the function f (x) whose domain is (0, positive infinity) satisfies that f (2x) = 2F (x) holds for any x ∈ (0, positive infinity)
We know that the function f (x) with the domain of (0, positive infinity) satisfies that for any x ∈ (0, positive infinity), f (2x) = 2F (x) holds, and when x ∈ (1,2], f (x) = 2-x, find the piecewise analytic expression of f (x)!

F (x) = 2 ^ (n + 1) - x x ∈ (2 ^ n, 2 ^ (n + 1)] n ∈ integer (both positive and negative)
Concrete solution
When x ∈ (2 ^ n, 2 ^ (n + 1)], n ∈ integer (both positive and negative)
Since f (2x) = 2F (x) x ∈ (2 ^ n, 2 ^ (n + 1)]
So f (x) = 2F (x / 2) x / 2 ∈ (2 ^ (n-1), 2 ^ n]
=2^2f(x/(2^2)) x/(2^2)∈（2^(n-2),2^(n-1)]
=2^3f(x/(2^3)) x/(2^3)∈（2^(n-3),2^(n-2)]
=……
=2^nf(x/(2^n)) x/(2^n)∈（2^0,2^1]=∈(1,2]
Because when x ∈ (1,2], f (x) = 2-x is substituted into the above formula
obtain
F (x) = 2 ^ NF (x / (2 ^ n)) = 2 ^ n [2-x / (2 ^ n)] = 2 ^ (n + 1) - x x ∈ (2 ^ n, 2 ^ (n + 1)] n ∈ integer (both positive and negative)

For a function f (x) whose domain is (0, positive infinity), f (2x) = 2F (x) holds for any x ∈ (0, positive infinity),
When x ∈ (1,2], f (x) = 2-x, the following conclusion is given?
1. The range of function f (x) is [0, positive infinity]
2. There exists n ∈ Z such that f (2 + 1) = 9
3. If K ∈ Z, (a, b) belongs to (k Power of 2, K + 1 power of 2), then the function f (x) increases monotonically in the interval (a, b)

It is easy to get that f (x) = 2F (x / 2) = 2 (2-x / 2) = 4-x, when x is located in (24], similar reasoning or mathematical induction can prove that f (x) = 2 ^ (n + 1) - x, when x is located in (2 ^ n, 2 ^ (n + 1)], where n can be a negative integer, that is, when n is an integer, the expression holds

When the definition field of X is (1,2), the function x ＾ 2 + MX + 4

When the definition field of X is (1,2), the function x ＾ 2 + MX + 44 or M

If the domain of function f (x2 + 2) is [0,2], find the domain of y = f (x)