Determine the domain and rage of the following functions F:R→R such that
1/√(9-x^2)
3x/(x-2)(x+1)
√((x)/(2-x))
X^2-1/√4x^3
The domain and range of each function are as follows:
1) f(x) = 1/√(9-x^2):
Domain: The function is defined for all real numbers except when the expression under the square root becomes negative. Therefore, the domain of f(x) is (-∞, -3] ∪ [3, ∞).
Range: Since the square root of a non-negative number is always non-negative, the range of f(x) is [0, ∞).
2) f(x) = (3x)/((x-2)(x+1)):
Domain: The function is defined for all real numbers except when the denominator equals zero. Therefore, the domain of f(x) is (-∞, -1) ∪ (-1, 2) ∪ (2, ∞).
Range: The function is a rational function and can take any real value except when it approaches a vertical asymptote. By analyzing the behavior of the function as x approaches the asymptotes, we can determine that the range of f(x) is (-∞, -∞) ∪ (-∞, ∞) ∪ (-∞, ∞).
3) f(x) = √(x/(2-x)):
Domain: The function is defined for all real numbers except when the expression under the square root becomes negative and when the denominator becomes zero. Therefore, the domain of f(x) is (-∞, 0) ∪ (0, 2).
Range: The square root of a non-negative number is always non-negative. Therefore, the range of f(x) is [0, ∞).
4) f(x) = (x^2-1)/√(4x^3):
Domain: The function is defined for all real numbers except when the expression under the square root becomes negative and when the denominator becomes zero. Therefore, the domain of f(x) is (-∞, 0) ∪ (0, ∞).
Range: The range of f(x) depends on the behavior of the expression as x approaches positive and negative infinity. If we approach positive infinity, the function tends to zero. If we approach negative infinity, the function approaches negative infinity. Therefore, the range of f(x) is (-∞, 0].
To determine the domain and range of the given functions, we need to consider any restrictions on the variables that would result in undefined values.
1. For the function f(x) = 1/√(9-x^2):
Domain: The square root of a negative number is undefined. Therefore, we need to ensure that 9-x^2 ≥ 0.
This condition simplifies to -3 ≤ x ≤ 3. Thus, the domain of the function is [-3, 3].
Range: Since the denominator of the function is a square root, the function can only output non-negative numbers. Thus, the range of the function is [0, ∞).
2. For the function f(x) = (3x)/((x-2)(x+1)):
Domain: We cannot have x-values that make the denominator equal to 0. Therefore, x ≠ 2 and x ≠ -1. Hence, the domain is all real numbers except x = 2 and x = -1.
Range: The range of this function is all real numbers because the function can take any value for valid inputs.
3. For the function f(x) = √(x/(2-x)):
Domain: Similar to the first function, the square root of a negative number is undefined. So, we need to ensure that x/(2-x) ≥ 0. We can solve this inequality by considering the signs of the numerator and denominator.
i) When x > 0, the denominator (2-x) will be negative, so the inequality is not satisfied. Thus, x > 0 does not satisfy the condition.
ii) When x < 0, both numerator and denominator are negative, so the inequality is satisfied.
iii) When x = 0, the denominator becomes undefined, so x = 0 is not included in the domain.
Therefore, the domain of the function is (-∞, 0).
Range: For x < 0, we have a square root of a positive number, so the range is (0, ∞).
4. For the function f(x) = (x^2 - 1)/√(4x^3):
Domain: The square root must be defined and the denominator must not be equal to zero.
To satisfy these conditions, we need to have 4x^3 > 0 and 4x^3 ≠ 1.
Solving 4x^3 > 0, we find that x ≠ 0.
Since 4x^3 ≠ 1, we find that x ≠ (1/4)^(1/3).
Hence, the domain of the function is (-∞, 0) ∪ (0, (1/4)^(1/3)) ∪ ((1/4)^(1/3), ∞).
Range: The range of this function depends on the values of x, but because of the complexity of the expression, it is difficult to determine a specific range without further evaluation.
Please note that the range of each function may vary depending on the values of x and the complexity of the expressions.