How to find a range (not domain) of function f(x)=square root of x^2-1/square root of 9-x^2
To find the range of the function f(x) = (sqrt(x^2 - 1)) / (sqrt(9 - x^2)), we can follow these steps:
Step 1: Start by determining the domain of the function. Since we have a square root in the numerator and denominator, we need to make sure that the expression inside each square root is non-negative.
For the numerator, x^2 - 1 must be greater than or equal to 0:
x^2 - 1 ≥ 0
Solving this inequality, we get:
(x - 1)(x + 1) ≥ 0
This means that x must be between -1 and 1 (inclusive) for the numerator to be non-negative.
For the denominator, 9 - x^2 must be greater than or equal to 0:
9 - x^2 ≥ 0
Solving this inequality, we get:
(x - 3)(x + 3) ≥ 0
This means that x must be between -3 and 3 (inclusive) for the denominator to be non-negative.
Therefore, the domain of the function is -3 ≤ x ≤ -1 or 1 ≤ x ≤ 3.
Step 2: Now, let's look at the behavior of the function as x approaches the boundaries of its domain. As x approaches -3 or -1 from the left, the numerator becomes increasingly negative, while the denominator remains positive. So the function approaches negative infinity.
As x approaches -3 or -1 from the right, the numerator becomes increasingly negative, while the denominator remains positive. Again, the function approaches negative infinity.
As x approaches 1 or 3 from the left, the numerator becomes increasingly positive, while the denominator remains positive. So the function approaches positive infinity.
As x approaches 1 or 3 from the right, the numerator becomes increasingly positive, while the denominator remains positive. Again, the function approaches positive infinity.
Therefore, the range of the function is (-∞, ∞) or all real numbers.
In summary, the range of the function f(x) = (sqrt(x^2 - 1)) / (sqrt(9 - x^2)) is (-∞, ∞) or all real numbers.