# how do i find the domain of this function?

### 4 Answers

- ?Lv 71 month agoFavourite answer
x + 6

f(x) = --------

x² - 4

// Look for restrictions: Since f(x) is a fraction,

// we can't have zero in the denominator, so

x² - 4 ≠ 0

x² ≠ 4

x ≠ ±2

DOMAIN of f(x) = all x ∊ ℝ such that x ≠ ±2,

written (-∞,-2) ∪ (-2,2) ∪ (2,+∞) .........ANS

// Your error is using the closed bracket at -∞

- 1 month ago
Think of the domain of any function as all the x values you can use. in this case, you wouldn't be able to use 2, or -2 because they would make the denominator undefined. so all numbers except -2 and 2.

- RaymondLv 71 month ago
The DOMAIN of a function is the set of values for which the function is "well-defined".

This includes the values for which the function gives one - and ONLY one - clear value.

We normally assume that the function's domain includes everything. Then we look for problem values (values for which the function is not clear).

One of the problems is when the function asks you to divide by zero.

Dividing by zero is "not defined". It does not matter that there eixsts case when it can be done. If the function asks you to divide by zero... then it is not "clearly defined".

Your function is

f(x) = (x+6) / (x^2 - 4)

when x = 2 AND when x = -2, the value of (x^2 - 4) is 0.

Therefore, when x = 2 and when x= -2, the function asks you to divide by zero.

For any other number, the function is "well defined" (meaning: it gives one, and only one, clear answer).

Your answer would be correct...

...except that you put a square bracket (implying inclusion) for the value called "infinity".

Infinity is NOT a well-defined value. That is why it is NEVER included.

It gets a ROUND bracket (nor a square one), because "infinity" itself is not included in "well-defined" values for the domain.

- PuzzlingLv 71 month ago
The domain is all real numbers other than -2 or 2, as you figured out. Your answer, in interval notation is correct, except for the [.

You can never reach -∞ or ∞, so they should *always* be shown with parentheses (open ended interval)

Answer:

(-∞, -2) ∪ (-2, 2) ∪ (2, ∞)

^