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).

step-by-step explanation:   divide the numerator by the denominator to get the decimal

it would still be 193

if divide by 100 then 1.93

i am not sure