# Domains of Rational Functions (standard notation)

Im preparing for a CLEP test in precalculus. As part of my prep, I need to review identifying domains of functions. I have a question about writing domains in standard notation. I was hoping someone could explain a bit the style.

For an example:

x-2 / x^2 -2x -35

As a rational expression, I know that the denominator can not be equal to zero. Therefore, to find the domain, I set the denominator equal to zero and solved the quadratic:

x = 7
x = -5

When x is either of these 2 values, the denominator will equal 0, and the expression is undefined. How would I write the domain in standard notation? I realize that the domain is all real numbers excluding -5 and 7, but is there a tidy way to write this?

Thanks.

symbolipoint
Homework Helper
Gold Member
{x | x<-5 U -5<x<7 U 7<x }

The domain is the union of the open intervals of x less than negative 5, x is greater negative 5 than but less than 7, and x is greater than 7.

Homework Helper
You could use interval notation.

$$(-\infty, -5) \cup (-5,7) \cup (7,\infty)$$

HallsofIvy
Homework Helper
{x | x<-5 U -5<x<7 U 7<x }

The domain is the union of the open intervals of x less than negative 5, x is greater negative 5 than but less than 7, and x is greater than 7.
I would NOT write it that way since the "U" notation is used for sets, not algebraic expressions. Either
$$\{x | x>-5 or -5<x< 7 or 7< x\}$$
or
$$\{x | x\ne -5 and x\ne 7\}$$
or
$$\{x| x< -5\}\cup \{x| -5< x< 7\}\cup \{x| x> 7\}$$

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Homework Helper
I've never seen

$$-5 < x > 7$$

considered a proper inequality: I believe Halls has a typo and that center piece
should be $\{x | -5 < x < 7 \}$.

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HallsofIvy
Homework Helper
Thanks, I have corrected it. (And will now pretend I never wrote such a silly thing!)

Mark44
Mentor
For an example:

x-2 / x^2 -2x -35
As a side note, an expression such as this written on a single line should be written with parentheses around the numerator and denominator, like so:
(x-2) / (x^2 -2x -35)

Under the order of operations, the expression as you wrote it would be interpreted to mean
x - (2/x2) - 2x - 35, which I'm sure isn't what you really meant.