Why is the domain in this problem all reals? (x^2-9)/(3x^2+1)

mmm4444bot said:
3x^2 + 1 does not factor (there's no factoring pattern for a 'sum of squares').

(3x + 1)(x - 1) = 3x^2 - 2x - 1

3x^2 + 1 is never zero; it's lowest value is 1. :cool:
Click to expand...


So only if you can't factor the denominator of the function then the domain is all reals right?
 
Ineedmathhelp123 said:
So only if you can't factor the denominator of the function then the domain is all reals right?
Click to expand...
No. The issue is not whether we can factor the numerator or denominator. The issue is that we cannot have zero in the denominator of a ratio. In other words, if any Real numbers for x lead to division by zero, then those numbers must be removed from the domain.

You need to determine whether there are any values of x which cause the denominator to equal zero.

3x^2 + 1 = 0

If you solve this equation, you will find that there are no Real solutions. Therefore, no Real number causes the denominator to equal zero. This means the domain contains all Real numbers.

Sometimes, we do factor a denominator, when trying to find values of x that cause it to equal zero. Yet, not all denominators factor (like in this exercise). Whenever we cannot factor one side of an equation set to zero, we need to solve it another way. :cool:
 
mmm4444bot said:
No. The issue is not whether we can factor the numerator or denominator. The issue is that we cannot have zero in the denominator of a ratio. In other words, if any Real numbers for x lead to division by zero, then those numbers must be removed from the domain.

You need to determine whether there are any values of x which cause the denominator to equal zero.

3x^2 + 1 = 0

If you solve this equation, you will find that there are no Real solutions. Therefore, no Real number causes the denominator to equal zero. This means the domain contains all Real numbers.

Sometimes, we do factor a denominator, when trying to find values of x that cause it to equal zero. Yet, not all denominators factor (like in this exercise). Whenever we cannot factor one side of an equation set to zero, we need to solve it another way. :cool:
Click to expand...


So if the problem can be solved when the denominator equals zero then its all reals except for the solutions (i.e. 4/X^2-3x-10 -> denominator factors to (x-5)(x+2) -> x=5,-2) so its all reals but 5 and -2 right.
 
Ineedmathhelp123 said:
4/(x^2 - 3x - 10) ->

denominator factors to (x - 5)(x + 2) ->

x = 5 or x = -2

so [the domain is] all reals but 5 and -2 right.
Click to expand...
Right.

x^2 - 3x - 10 = 0

You found the Real solutions, by factoring. That's good, when the denominator factors. If it does not factor, then you need to find another way, like with this one:

4/(x^2 - 3x - 11)
 
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