Determining the Natural Domain of Functions: True or False?

Determining the Natural Domain of Functions: True or False?Let's analyze each statement and determine its validity:(1) The natural domain of f(x) = (√(4 - x^2))/(x^2 - x - 6) is [-2, 2].**False. While the square root restricts the domain to -2 ≤ x ≤ 2, we must also consider the denominator. The denominator, x^2 - x - 6, cannot equal zero. Factoring, we get (x - 3)(x + 2) ≠ 0. Therefore, x ≠ 3 and x ≠ -2. The natural domain of f(x) is [-2, 2] / {-2, 3}, which means all values between -2 and 2, excluding -2 and 3.(2) The natural domain of G(y) = √((y + 1)^-1) is {y ∈ R: y > -1}.**True. The expression inside the square root must be non-negative. Since (y + 1)^-1 is the same as 1/(y + 1), the denominator cannot be zero, and the entire expression is only defined for y > -1.(3) The natural domain of h(u) = arcsin(2u + 3) is (-∞, ∞).**False. The arcsine function (arcsin) has a restricted domain of [-1, 1]. This means the input, 2u + 3, must fall within this range:-1 ≤ 2u + 3 ≤ 1Solving this inequality, we get -2 ≤ u ≤ -1. The natural domain of h(u) is [-2, -1].(4) The natural domain of F(t) = t^(3/2) - 4 is {t ∈ R: t ≥ 0}.**True. The expression t^(3/2) involves a square root (t^(1/2)). Since the square root is only defined for non-negative values, t must be greater than or equal to zero.**In summary:**Understanding the restrictions imposed by square roots, denominators, and inverse trigonometric functions is crucial for accurately determining the natural domain of a function.