find the domain and range of:
the square root of 25-x^2
f(x)=√(25-x²)
For f(x) to be real, we require that the expression 25-x² be non-negative, i.e.
25-x²≥0 ....(1)
(1) is satisfied when
|x|≤5,
or x∈[-5,5] which is exactly the domain of f(x).
The range of f(x) varies from 0 when x=±5 to 5 when x=0.
To find the domain and range of the function f(x) = √(25 - x^2), we need to consider two things: the domain restrictions and the possible output values.
1. Domain:
The square root of a negative number is not defined in the real number system. Therefore, in order for f(x) to be defined, the expression 25 - x^2 must be non-negative. This means that:
25 - x^2 ≥ 0
To solve this inequality, we can factor it as:
(5 - x)(5 + x) ≥ 0
From this, we can see that x must satisfy:
5 - x ≥ 0 and 5 + x ≥ 0
Solving for x in each inequality:
5 - x ≥ 0 ➜ x ≤ 5
5 + x ≥ 0 ➜ -5 ≤ x
Therefore, the domain of the function is:
-5 ≤ x ≤ 5
2. Range:
Since the square root of a non-negative number is a real number, the range of the function f(x) is all non-negative real numbers, or [0, ∞).
In summary:
Domain: -5 ≤ x ≤ 5
Range: [0, ∞)
To find the domain and range of the function f(x) = √(25 - x^2), we need to consider the restrictions on the input (x) and output (f(x)) of the function.
1. Domain:
The domain represents all possible values that x can take in the function. In this case, the expression inside the square root (25 - x^2) must be non-negative, as the square root of a negative number is undefined.
So, we set 25 - x^2 ≥ 0 and solve for x:
25 - x^2 ≥ 0
x^2 - 25 ≤ 0
(x - 5)(x + 5) ≤ 0
The solutions to this inequality are -5 ≤ x ≤ 5. Therefore, the domain of the function is -5 ≤ x ≤ 5.
2. Range:
The range represents all the possible values that f(x) can take as x varies in the defined domain. Since the square root of a non-negative number is always non-negative, there are no restrictions on the output (f(x)).
Therefore, the range of the function is f(x) ≥ 0.
In summary:
Domain: -5 ≤ x ≤ 5
Range: f(x) ≥ 0