Domain of g(x)
g(x)= sqrt(3-(sqrt(25-x^2))
I got -5, -4, 4, and 5.
I wrote it like [-5, -4] U [4, 5]
I don't think that's right?
Under the overall square root you have
3 - square root of 25-x^2
What is under the radical sign has to be positive.
Domain is the X values that will result in a solution and not undefined.
It looks like you can have x=-5 and x =5 because that will give a square root of 0 to subtract from 3. That works. I see that your -4 and 4 also will work.
Those values are included and you used the [ ] instead of ( ) so that is correct and you used the U and that is correct because both sets of the values work. You wouldn't want an intersection.
I agree with your answer.
It could also be written as
-5 =<x=< -4 or 4 =< x =<5
but most likely your teacher wants interval notation whihc your provided.
Thank you!
You are welcome!
To find the domain of the function g(x) = sqrt(3 - sqrt(25 - x^2)), we need to consider the restrictions on the input values (x) that make the expression inside the square root valid.
First, let's analyze the inner square root: 25 - x^2. For the expression inside the square root to be valid, it must be greater than or equal to zero.
25 - x^2 ≥ 0
To solve this inequality, we can factor it as:
(x - 5)(x + 5) ≥ 0
Now, we have two critical points: x = -5 and x = 5. These points divide the number line into three intervals: (-∞,-5], [-5, 5], and [5, ∞).
We need to test a value from each interval to determine whether the inequality is true or false.
- Let's test x = -6, which is in the interval (-∞,-5]:
( -6 - 5)( -6 + 5) ≥ 0
-11( -1) ≥ 0
11 ≥ 0 → True
- Let's test x = 0, which is in the interval [-5, 5]:
(0 - 5)(0 + 5) ≥ 0
-5(5) ≥ 0
-25 ≥ 0 → False
- Let's test x = 6, which is in the interval [5, ∞):
(6 - 5)(6 + 5) ≥ 0
1(11) ≥ 0
11 ≥ 0 → True
From the tests, we can conclude that the valid intervals are (-∞,-5] and [5, ∞), as they satisfy the inequality. The interval [-5, 5] is excluded since it does not satisfy the inequality.
Therefore, the correct domain for the function g(x) = sqrt(3 - sqrt(25 - x^2)) is:
(-∞,-5] U [5,∞)