a)

b)

Need help

c)

d)

e)

2. Re: Finding the domain - Need correcting please

Originally Posted by YassAttack

a)

b)

Need help

c)

d)

e)

a) Incorrect, it's defined everywhere except where the denominator is 0. Where is this?

b) Defined wherever the stuff under the square root is nonnegative. Where is this? If you need help solving $\displaystyle x^2 - 4x + 2 \geq 0$, start by completing the square...

c) Defined wherever the denominator is nonzero and wherever the stuff under the square root is nonnegative. Where is this?

d) Defined wherever the denominator is nonzero. Since the denominator will only be 0 if |y| = -3, which is impossible, it's defined over R, like you wrote, so correct

e) Incorrect, it's defined wherever $\displaystyle |z - 3| - 1 \geq 0$. Where is this?

3. Re: Finding the domain - Need correcting please

I think I just didn't write my answers properly...
like for a) i meant to write all real numbers except for -1 and 1.
c) all real numbers except 4 and -2
e) -(infinity, +infinity) since whatever number in the |abs| will become positive. so |x| - 1 > 0.

Did i just mess up in the way i wrote that?

and b can't be factored.

4. Re: Finding the domain - Need correcting please

Originally Posted by YassAttack
I think I just didn't write my answers properly...
like for a) i meant to write all real numbers except for -1 and 1.
c) all real numbers except 4 and -2
e) all real numbers except for 3

Did i just mess up in the way i wrote that?

and b can't be factored.
In that case a) is correct. Write it as $\displaystyle \mathbf{R}\backslash\{-1, 1\}$ or $\displaystyle (-\infty, -1)\cup (-1, 1)\cup (1,\infty)$.

I'm pretty sure b) CAN be factored... It might not factorise over the rationals though. Like I said, try completing the square...

c) I doubt that is the answer, as you need to take more into account than that. The square root function is only defined when the stuff under it is nonnegative. For that stuff under the square root to be 0, then the numerator (x) is 0, which is fine. For the stuff to be positive, then the top and bottom need to both be positive or both be negative. When does this happen?

e) Not even close. Do you know how to solve inequalities that involve absolute values?

5. Re: Finding the domain - Need correcting please

Originally Posted by Prove It
In that case a) is correct. Write it as $\displaystyle \mathbf{R}\backslash\{-1, 1\}$ or $\displaystyle (-\infty, -1)\cup (-1, 1)\cup (1,\infty)$.

I'm pretty sure b) CAN be factored... It might not factorise over the rationals though. Like I said, try completing the square...

c) I doubt that is the answer, as you need to take more into account than that. The square root function is only defined when the stuff under it is nonnegative. For that stuff under the square root to be 0, then the numerator (x) is 0, which is fine. For the stuff to be positive, then the top and bottom need to both be positive or both be negative. When does this happen?

e) Not even close. Do you know how to solve inequalities that involve absolute values?
Oh man you're a genius.

b) I completed the square and got x = 3.414 and x = -0.586
so (-∞, -0.586) (3.414, ∞) is what I got?

c) (4, ∞)

e) not a clue...can you give me a tip?

6. Re: Finding the domain - Need correcting please

Originally Posted by YassAttack
Oh man you're a genius.

b) I completed the square and got x = 3.414 and x = -0.586
so (-∞, -0.586) (3.414, ∞) is what I got?

c) (4, ∞)

e) not a clue...can you give me a tip?
b) Keep your answers exact. I suspect you also didn't really know what to do when you got to an absolute value in your inequality...

You should know that an absolute value represents the distance between that number and the origin. So |x| = a means that x = -a or x = a (since both of those are a units away from the origin).

What happens if you have $\displaystyle |x| < a$? It means that the x can take on any value that has a distance that's smaller than $\displaystyle a$ units away from the origin.
So $\displaystyle -a < x < a$.

What happens if you have $\displaystyle |x| > a$? It means that the x can take on any value that has a distance that's greater than $\displaystyle a$ units away from the origin. Therefore $\displaystyle x < -a$ or $\displaystyle x > a$.

So if you have $\displaystyle |z - 3| - 1 \geq 0$, then $\displaystyle |z - 3| \geq 1$. In other words, the size of $\displaystyle z - 3$ can be anything greater than or equal to 1. Does it make sense then to say that $\displaystyle z - 3 \leq -1$ or $\displaystyle z - 3 \geq 1$?
What values therefore can $\displaystyle z$ take?

7. Re: Finding the domain - Need correcting please

Originally Posted by Prove It
b) Keep your answers exact. I suspect you also didn't really know what to do when you got to an absolute value in your inequality...

You should know that an absolute value represents the distance between that number and the origin. So |x| = a means that x = -a or x = a (since both of those are a units away from the origin).

What happens if you have $\displaystyle |x| < a$? It means that the x can take on any value that has a distance that's smaller than $\displaystyle a$ units away from the origin.
So $\displaystyle -a < x < a$.

What happens if you have $\displaystyle |x| > a$? It means that the x can take on any value that has a distance that's greater than $\displaystyle a$ units away from the origin. Therefore $\displaystyle x < -a$ or $\displaystyle x > a$.

So if you have $\displaystyle |z - 3| - 1 \geq 0$, then $\displaystyle |z - 3| \geq 1$. In other words, the size of $\displaystyle z - 3$ can be anything greater than or equal to 1. Does it make sense then to say that $\displaystyle z - 3 \leq -1$ or $\displaystyle z - 3 \geq 1$?
What values therefore can $\displaystyle z$ take?
(4, ∞)?

But wouldn't any number you put in the |abs| become positive automatically? I don't understand why any number wouldn't be okay then except for 3 because |3-3|-1 = -1