Help with pre-cal problems?

Question

how do I find the domian, range and zeros of this function: f(x)= Sq. root of x^2 - 16??

Thanks for the help!

Answer 1

Sq. root of x^2 - 16
If -4 < x < 4 then f(x) is imaginary
So Domain is (-infinity,-4] U [4, +infinity)

The range is y>= 0

The x-intercepts are (-4,0) and (4,0)

Answer 2

first of all. The domain are all possible values for x. As you have a square root, you need that your expression has a positive sign or zero.

I think your expression is f(x) = sqrt (x^2 -16) so you need
x^2 - 16 >= 0

In order to solve this inequality you need to find the zeros or the expression:

x^2 - 16 = (x+4) (x-4) so there are change of sign at x=-4 and x = 4.

So I have 3 regions:

region 1 from -infinite to -4
region 2 from -4 to 4
region 3 from 4 to + infinite

We check signs in each region by taking any value.

region 1 : I choose x = -5 and I substitute: (-5-4)(-5+4) and the sign is positive, so it is a valid region.
region 2: I choose x = 0 and I substitue: (0 -4) (0+4) and the sign is negative, so it is not a valid region.
region 3. I shoose x = 6 and I substitute: (6-4) (6 + 4) and the sigh is positive, so it is a valid region.

In few words;: the domain is { x | x E (-infinite, -4) U (4, +infinite).

MariLuz

Answer 3

SQRT(x^2 - 16) exists only if x^2 - 16 >=0, which means x^2>=16, which means:

x>=+4 or x<=-4. This is the domain.

The range is all non-negative real numbers.