# Math Help - Domain Question

1. ## Domain Question

How do I find the Domain of

x+4 / square root of 3x+1?

Square root of 2x+5?

X / X(X-3)?

I have a few more questions on the bottom. Thanks

2. #1. What values of x will result in a 0 in the denominator or a negative inside the radical?. Those are NOT in the domain.

3. Originally Posted by galactus
#1. What values of x will result in a 0 in the denominator or a negative inside the radical?. Those are NOT in the domain.
No idea Is there a way or method anyone can tell me so I can just solve it? Thanks

4. A square root is defined for nonnegative real numbers and a fraction is defined when the denominator is $\neq$0.

5. You have $\sqrt{3x+1}$ in the denominator. You can not have division by 0, therefore, set $\sqrt{3x+1}=0$ and solve for x.

Also, find what values then give a negative value inside the radical. That should be easy. Yes?.

6. Originally Posted by galactus
You have $\sqrt{3x+1}$ in the denominator. You can not have division by 0, therefore, set $\sqrt{3x+1}=0$
This is not correct.

You can't set $\sqrt{3x+1}=0$ 'cause you'll have a denominator with zero.

It should be $3x+1>0$, which is easy to solve.

7. Originally Posted by Krizalid
This is not correct.

You can't set $\sqrt{3x+1}=0$ 'cause you'll have a denominator with zero.

It should be $3x+1>0$, which is easy to solve.

So X > -1/3 or (-1/3, Infinite)?

8. Originally Posted by JonathanEyoon
So X > -1/3 or (-1/3, Infinite)?
correct

remember, the domain of a function is the set of all inputs (in this case, x-values) for which the function is defined. We see that the function is undefined if the denominator is zero, so we must have $\sqrt {3x + 1} \neq 0$. But there is also a restriction on the square root, what is being square rooted cannot be negative. The only way we can satisfy both these conditions is if $3x + 1 > 0$. Thus the domain is all x such that $3x + 1 > 0$ or, more formally,

(Using set notation) $dom(f) = \{ x \mbox { : } 3x + 1 > 0 \} = \{ x \mbox { : } x > 1/3 \}$

(Using interval notation) $dom(f) = \left( - \frac {1}{3} , \infty \right)$

9. My point was, setting it equal to 0 and solving for x would result in -1/3, which is not in the domain. You made a better point.

10. Thanks guys this helped a bunch~ I'm currently learning operations and compositions of functions and I have a few questions I was wondering you could answer for me.

1. f(x) = x^2, g(x) = squareroot 1-x
I'm trying to find (f o g)(x) and it's domain. The problem i'm having is I know that (f o g)(x) = 1 - x so shouldn't the Domain be (-Infinite, Infinite)?

2. Similarily same functions but finding (g o f)(x). It is = Squareroot 1 - x^2. So to find the Domain, I should use 1 - x^2 greater than equal to 0?

In both these cases i'm having trouble understanding what the Domain should be. If anyone can help explain it would be great.

Oh and for one of the above questions I had earlier, Squareroot 2x + 7 domain should be 2x+7 Greater than equal to 0 where it should end up x greater than equal to [-7/2, infinite) right?

11. You really should post new questions in new threads
Originally Posted by JonathanEyoon
Thanks guys this helped a bunch~ I'm currently learning operations and compositions of functions and I have a few questions I was wondering you could answer for me.

1. f(x) = x^2, g(x) = squareroot 1-x
I'm trying to find (f o g)(x) and it's domain. The problem i'm having is I know that (f o g)(x) = 1 - x so shouldn't the Domain be (-Infinite, Infinite)?
$(f \circ g)(x) = |1 - x|$ and yes, this is defined for all real x, so the domain is as you say

12. Originally Posted by JonathanEyoon
2. Similarily same functions but finding (g o f)(x). It is = Squareroot 1 - x^2. So to find the Domain, I should use 1 - x^2 greater than equal to 0?

In both these cases i'm having trouble understanding what the Domain should be. If anyone can help explain it would be great.
recall my explanation of what the domain is in one of my previous posts in this thread. it is the set of all x's for which the function is defined.

you are correct in that $(g \circ f)(x) = \sqrt {1 - x^2}$

and you are correct that the domain is all x such that $1 - x^2 \geq 0$ since the square root is only defined for these values.

now to solve, which i don't see what your problem is. we have the difference of two squares on the left

$1 - x^2 \geq 0$

$\Rightarrow (1 + x)(1 - x) \geq 0$

$\Rightarrow -1 \leq x \leq 1$

thus, the domain is $[-1,1]$

13. Originally Posted by JonathanEyoon
Oh and for one of the above questions I had earlier, Squareroot 2x + 7 domain should be 2x+7 Greater than equal to 0 where it should end up x greater than equal to [-7/2, infinite) right?
you typed $\sqrt {2x + 5}$ before. which is it? anyway, the domain for the function you found above is correct. except you don't say "x greater than equal to [-7/2, infinite)" that statement makes no sense! it is x >= -7/2 or dom(f) = [-7/2, infinity)