Answer:
1) (-∞, -2) ∪ (2, ∞); 2) (-4, 4)
Step-by-step explanation:
1)
f(x)+root x^2-4 => f(x) = √(x^2 - 4)
Since the domain of the square root function is [0, ∞), we must determine the intervals on which x^2 - 4 is ≥ 0. First determine the x values at which x^2 - 4 = 0: (x-2)(x+2) = 0 produces two results: x = -2 and x = 2.
These two numbers determine three intervals on the real number line:
(-∞, -2), (-2, 2), (2, ∞). Choosing a "test number" from each interval, we get -3, 0 and 3. All that remains to do now is to determine whether x^2 - 4 is positive or negative on each interval.
Case 1: test number -3: Is (-3)^2 - 4 positive or neg? It's positive.
Case 2: test number 0. Is (0)^2 -4 positive or neg? It's negative. Reject this interval.
Case 3: test number +3: Is (+3)^2 - 4 positive or neg? It's positive.
Thus, f(x) = √(x^2 - 4) is defined on (-∞, -2) ∪ (2, ∞); this is the domain.
2) f(x) = √(16 - x^2). Domain? We go thru steps similar to those above. 16 - x^2 factors into (4 - x)(4 + x); the roots are 4 and -4, and the number intervals are (-∞, -4), (-4, 4), (4, ∞)
Case 1: Is the radicand (x^2 - 4) + or - on (-∞, -4)?
Choose the test number -6 from the interval (-∞, -4). (16 - x^2) is negative on this interval, so (-∞, -4) is NOT part of the domain of f(x) = √(16 - x^2). Next, choose the test number 0; you'll find that f(x) = √(16 - x^2). is positive, so (-4, 4) IS part of the domain of f(x) = √(16 - x^2). Choosing the test number +6, you'll find that (16 - x^2) is negative, so (4, ∞) is NOT part of the domain of f(x) = √(16 - x^2).
Conclusion: the domain of f(x) = √(16 - x^2) is (-4, 4).
