Find the domain and range of the following functions : 1) f (x) + root x^2-4 2) f (x) + root 16-x^2

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).