Find the definition domain of function: F (x) = (radical x + 4) / x + 2

Find the definition domain of function: F (x) = (radical x + 4) / x + 2

∵ f (x) = (root x + 4) / x + 2
ν x + 4 ≥ 0, so x ≥ - 4
x≠0
So the definition domain of the original function is [- 4,0) ∩ (0, ∞)
-------------Hope to adopt
In addition, f (x) = (radical x + 4) / x + 2 and f (x) = (radical x + 4) / (x + 2)
It's a different function. Do it yourself

Let f (x + 1) = x + 2, find f (x), f (x + 1), f (the square of x) 2. If f (1-x / 1 + x) = 1-x square / 1 + x square, then the analytic formula of F (x) is? 3. Given that f (1 = 1 / x) = 1 / X squared - 1, find f (x) 4. Given that the quadratic function satisfies f (3x + 1) = 9x square - 6x + 5, find f (x) Sorry, the number 3 is wrong. It should be 3. Given that f (1 + 1 / x) = 1 / X squared - 1, find f (x) The computer doesn't know how to make the square root, so it's hard to understand

1. F (x) = x ^ 2-1, f (x + 1) = x ^ 2 + 2x, f (x ^ 2) = x ^ 4-1f (radical x + 1) is defined as x > = 0, f (x) is [1, + ∞), f (x + 1) is [0, + ∞), f (x ^ 2) is (- ∞, - 1] u [1, + ∞) 2, f (x) = - 3x ^ 2-4x-1) / (5x ^ 2 + 4x + 1) 3,? 4, f (x) = x ^ 2-4x + 8

The definition domain of (x + 6 / 1) - 1 under the radical sign f (x) = is set, and the domain of a function g (x) = LG (- x ^ 2 + 2x + m) is set Finding the intersection of complement a and B when m = 3 If a intersects B = (- 1,4), find the real number M

F (x) = radical {(5-x) / (x + 1)}
∴｛（5-x）/（x+1）｝＞=0
ν 5-x ＞ = 0 x + 1 ＞ 0 (∵ x + 1 in denominator, ᙽ 0)
x＜=5 x＞-1
∴A=(-1,5]
When G (x) = LG (- x ^ 2 + 2x + m) M = 3
-x^2+2x+3＞0
（x-3）（x+1）＜0
∴-1＜x＜3
B=（-1,3）
A∩B=（-1,3）
∵A∩B= （-1,4） 5＞4
/ / 4 is a solution of - x ^ 2 + 2x + M
By introducing 4 into the equation, we get: - 16 + 8 + M = 0
M=8
Thank you for your adoption

Given the complete set u = R, function y = X-2 + the definition domain of X + 1 under the root sign is a, and the definition domain of 2x + 4 under the root sign of function y = x + 3 is B, then find the set a, B

If the definition domain of the function y = X-2 + X + 1 is a, then a is the solution set of X-2 > = 0 and X + 1 > = 0, so a is {x | x > = 2}
If the definition field of 2x + 4 under the root sign of function y = x + 3 is B, then B is 2x + 4 > = 0 and X + 3 is not equal to 0, so B is {x | x > = - 2}

Let the complete set u = R, set a be the domain of function f (x) = radical 3-x + radical x + 1, and set B be the domain of function g (x) = root 4-x of root X Then the question above is C = {x| x < a} Find a ∩ B a ∪ B If B ∩ C = B, find the value range of real number a If a ∩ C ≠ ∩ to find the value range of A

According to my understanding of the title:
(1) First calculate the set a and B
A=[-1,3] B=(0,4] A∩B==[0,3] A∪B=[-1,4]
(2) If B ∩ C = B, B is a subset of C, C = {x | x < a} so a > 4 (note that it cannot be equal to 4)
(3) Draw the number axis, the end point of C is a a ∩ C ≠ ∩ as long as a > - 1
Contact with questions!

Let's ask you to know that the definition field of the function f (x) = 1 / 1 of the radical 3-x + (root x + 2) is set a, B = {x | x is less than a} The radical of (x) is defined as the radical of (x) = 1

If a is included in B, a is a subset of B. If a = - 1, CRA = {x | x < = - 2 or x > 3} CRB = {x | x > 3} a ∩ (CRB) = {x | x > 3} CRB = {x | x > = - 1} a ∩ (CRB) = {x | x > 3}

The radical of (x) is defined as the radical of (x) = 1 If the complete set u = {x | ≤ 4}, a = - 1, find CUA and a ∩ (cub)

From the meaning of the title
3-x≧0,x﹥0,
The solution is 0 ﹤ x ≤ 3 ﹤ a = {x | 0 ﹤ x ≤ 3}
∵ A is contained in B, ᙽ a ≥ 3
(2) ∵ a = - 1, u = {x ∵ x ≤ 4}
ν B = {x ≤ - 1}
CuA=3＜x≦4
A∩(CuB)=A=0＜x≦3

The definition domain of (x-x ^ 2) under the radical of function f (x) = is Why did the teacher criticize me for being wrong

Open mode must be ≥ 0
So x-x 2 ≥ 0,
x²－x≦0.
0≦x≦1.
A: the definition domain is
0 ≤ x ≤ 1. Or write as [0, 1]

Given the function f (x) = X-1 under the radical sign, then the definition domain of function y = f [(1 / 2) x]

0.5^x≥1
x≥0
The definition domain of function y = f [0.5 ^ x] is x ≥ 0

The definition domain of the function f (x) = (1-x * 2) + (x * 2-1) is

1-x2 > = 0, x2-1 > = 0, domain {- 1,1}