# Math Help - [SOLVED] Find the domain of the function

1. ## [SOLVED] Find the domain of the function

f(x) = log(x^2 - 2x - 3)

2. Originally Posted by moonman
f(x) = log(x^2 - 2x - 3)
Find the solution set for $x^2 -2x -3 >0$.

3. Originally Posted by Plato
Find the solution set for $x^2 -2x -3 >0$.
x^2 -2x -3 > 0

(x+1)(x-3)
x=-1 x=3 is this the answer?

4. Originally Posted by moonman
x^2 -2x -3 > 0

(x+1)(x-3)
x=-1 x=3 is this the answer?
You have solved x^2 - 2x - 3 = 0. Is that what Plato instructed you to solve ....?

5. Originally Posted by mr fantastic
You have solved x^2 - 2x - 3 = 0. Is that what Plato instructed you to solve ....?
Yes?

6. Originally Posted by moonman
Yes?
i recall seeing a ">" sign in Plato's post, as opposed to an "=" sign

7. Originally Posted by Jhevon
i recall seeing a ">" sign in Plato's post, as opposed to an "=" sign
ahhh~!!!
How do I solve when those > < things are there?

8. Originally Posted by moonman
ahhh~!!!
How do I solve when those > < things are there?
Have you been taught anything about solving inequations? Have you seen equations like this before?

9. Originally Posted by mr fantastic
Have you been taught anything about solving inequations? Have you seen equations like this before?
I've been away from math for so long
I know that > means greater than and this < means less than.
The question has to do with domain, so is the domain -1 to 3?

10. Originally Posted by moonman
ahhh~!!!
How do I solve when those > < things are there?
ok, what you did is a start, we found where it IS zero, we want to find where it's bigger. so we can check the intervals that our solution breaks up the real line into

we have x = -1 and x = 3

so draw a number line and mark this on it. now, test numbers on both ends of the number line, and inbetween the numbers. if you get a positive answer, then that region is where we are greater than zero, if you get a negative number, that's where it's less than zero

incidentally, domain is just the set of x-values for which a function is defined. we have to solve for where this expression is positive, because the logarithm is only defined where its argument is positive. so once you answer the above question, that's your domain

11. Originally Posted by Jhevon
ok, what you did is a start, we found where it IS zero, we want to find where it's bigger. so we can check the intervals that our solution breaks up the real line into

we have x = -1 and x = 3

so draw a number line and mark this on it. now, test numbers on both ends of the number line, and inbetween the numbers. if you get a positive answer, then that region is where we are greater than zero, if you get a negative number, that's where it's less than zero

incidentally, domain is just the set of x-values for which a function is defined. we have to solve for where this expression is positive, because the logarithm is only defined where its argument is positive. so once you answer the above question, that's your domain

How do I test the numbers?

12. Originally Posted by moonman
How do I test the numbers?
plug them into the original quadratic, (x + 1)(x - 3), evaluate and see if your answer is positive or negative

13. Originally Posted by Jhevon
plug them into the original quadratic, (x + 1)(x - 3), evaluate and see if your answer is positive or negative

Okay, so I plugged in a -3:
(-3 + 1) (-3 - 3)
so that gave me positive 12

Then I plugged in a -1
(-1 + 1) (-1 - 3)
so that gave me neutral 0

Then I plugged in a 1
(1 + 1) (1 - 3)
so that gave me negative 4

Then I plugged in a 3
(3 + 1)(3 - 3)
so that gave me neutral 0

Then I plugged in a 5
(5 + 1)(5 - 3)
so that gave me positive 12

Did i do this right? What's next?

14. Originally Posted by moonman
Okay, so I plugged in a -3:
(-3 + 1) (-3 - 3)
so that gave me positive 12

Then I plugged in a -1
(-1 + 1) (-1 - 3)
so that gave me neutral 0

Then I plugged in a 1
(1 + 1) (1 - 3)
so that gave me negative 4

Then I plugged in a 3
(3 + 1)(3 - 3)
so that gave me neutral 0

Then I plugged in a 5
(5 + 1)(5 - 3)
so that gave me positive 12

Did i do this right? What's next?
There are big gaps in your understanding. Start filling them in by reading this: Solving Quadratic Inequalities: Concepts

15. Domain is all real numbers? ?

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