Domains of Rational Functions (standard notation)

[Shafty]

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]

{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.

 

[statdad]

You could use interval notation.

<br /> (-\infty, -5) \cup (-5,7) \cup (7,\infty)<br />

 

[HallsofIvy]

{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&gt;-5 or -5&lt;x&lt; 7 or 7&lt; x\}
or
\{x | x\ne -5 and x\ne 7\}
or
\{x| x&lt; -5\}\cup \{x| -5&lt; x&lt; 7\}\cup \{x| x&gt; 7\}

 

[statdad]

I've never seen

<br /> -5 &lt; x &gt; 7<br />

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

 

[HallsofIvy]

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

 

[Mark44]

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.