How do you find the domain of #f(x)=(sqrt(x-9))/(x^2-16)#?

1 Answer
NickTheTurtle · Daniel L.
Apr 2, 2017

#{x|x≥9}# or #[9, oo[#

Explanation:

The domain is the set of numbers that #x# can be equal to.

Let's look at the numerator first. You can only take the square root of a number greater than or equal to #0#. That means that #x-9≥0# (else you will be taking the square root of a negative number). Solving this gives #x≥9#.

Let's then look at the denominator. You cannot divide by zero. This means that #x^2-16ne0#. Solving this gives #xne4vvxne-4#.

Combining these gives #x≥9 ^^ xne4 ^^ xne-4#. However, since #x≥9# implies that #xne-4# and #xne4#, we can remove both.

Our final answer is #x≥9#. In set notation, we write #{x|x≥9}#. In interval notation, we write #[9, oo[#.

Related questions
See all questions in Domain and Range of a Function
Impact of this question
2305 views around the world
You can reuse this answer
Creative Commons License