Pre calc question? Stuck?

Question

Give the domain, range, and zeros of the function f(x)= the square root of x^2 -16. Thanks in advanced for the help.

Answer 1

Domain is what numbers x work in f (x).
We can't take the square root of a negative number so

x^2 - 16 >= 0

x^2 >= 16

x >= 4 or x <= -4

Domain: (-inf , -4 ] U [ 4 , +inf)

Range is the output of the f (x) for all numbers of the domain.

Range: [ 0 , +inf)

Zeros are wher f (x) = 0, so set the function to zero

sqr ( x ^ 2 - 16) = 0, then square both sides

x ^ 2 - 16 = 0

x ^ 2 = 16

x = + and - 16

Answer 2

This is a difference of squares, so you can factor it as:
sqrt[(x - 4)(x + 4)]

The zeroes are x = 4, x = -4

The domain (assuming you can't take the square root of negative numbers) is x <= -4 or x => 4

The range (again assuming you only want the positive square roots) is y >= 0.

Answer 3

The domain is all real x less than -4 or greater than 4, inclusive. The range is all real y greater than or equal to 0. The zeros of f(x) are -4 and 4.

Answer 4

well, x^2-16 has to be greater or equal to 0, so x=<-4 or x=>(or equal to) 4. (that's the domain)

range would be: 0 to positive infinity

zeroes: x = -4, 4

Answer 5

This is just a note. You have square root of x^2 -16. Some answers have interpreted them as sqrt(x^2-16) while others have interpreted it as sqrt(x^2) -16. You may wish to add an additional comment.

Answer 6

Use the guidelines: Cos(x)= Adj/Hyp Sin(x)=Opp/Hyp Cot(x)=a million/tan(x) which equals: cos(x)/sin(x) Draw the triangle, then label the regular values (3 and four) on the right factors. Then, use the Pythagorean Theorem to locate the lacking component. After that, merely persist with the guidelines