# Pre calc review - interval notation

1. Aug 26, 2014

### datafiend

1. The problem statement, all variables and given/known data
What's the domain of 14/x2-x-6, in interval notation?

2. Relevant equations

$\underline{14}$x2-x-6

3. The attempt at a solution
[3, + infinity)
[-2,- infinity)
Sorry, I can't find the infinity symbol

Thx
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 26, 2014

### Ray Vickson

You wrote
$$\frac{14}{x^2} -x-6$$
when read using standard rules for parsing mathematical expressions. Did you really mean that, or did you want
$$\frac{14}{x^2 - x - 6} \: ?$$
If you meant the latter you absolutely MUST use parentheses, like this: 14/(x^2 - x - 6).

3. Aug 26, 2014

### datafiend

My mistake, the denominator SHOULD be in parentheses.

4. Aug 26, 2014

### LCKurtz

What about values between 3 and -2?

5. Aug 26, 2014

### HallsofIvy

Staff Emeritus
You mean 14/(x2- x- 6)= 14/((x- 3)(x+ 2))

No, this is wrong because it does not include such numbers as x= 0 for which 14/(0- 0- 6)= -7/3 or x= -5 for which 14/(25+ 5- 6)= 7/12.

Strictly speaking this is bad notation- it should be (-infinity, -2]. (The smaller goes on the left.)
But even that is not correct because it does not contain x= 0 and x= 5 for which 14/(25- 5- 6)= 1.

The correct answer is NOT a single interval, it is a union of three disjoint intervals.

6. Aug 27, 2014

### skiller

I'm surprised no one's mentioned the incorrect use of the square bracket to end the set.