Finding the domain of X write in interval notation
f(x)= 4x+2 sq.

- anonymous

Finding the domain of X write in interval notation
f(x)= 4x+2 sq.

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- amistre64

since "x" is only apart of 4x, there are no restrictions to it, the domain of f(x) is all real numbers: (-inf,inf)

- amistre64

I noticed from another post, that you might mean that all this is under the radical sign...
sqrt(4x+2) ??

- anonymous

yes

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## More answers

- amistre64

it is good to use the paranthesis to indicate inclusion :)
since "even" roots care about the sign of the answers, this square root doesnt want anything that turns it negative.
We restrict our "domain" to ll real xs that keep this a zero or positive number.

- anonymous

how do i solve it?

- amistre64

You would set the 4x+2 = 0 and solve for x
4x+2 = 0
4x = -2
x = -2/4
x = -1/2
since this number makes everything under the radical equal to zero, any number less than this will produce a negative result and that is bad. Every number ablve this value will stay positive,
to the domain is:
x such that x is an element of the set [-1/2, inf)

- anonymous

so if I had f(x)=\[\sqrt{6-3x}\] i would add 6 to both sides

- amistre64

you would notice first of all that this is a square root problem, and that square roots are forbidden to have any negative value under their umbrella.
So we equate the problem that is under the radical sign to zero.
6 - 3x = 0
we need this equation here to be either zero or positive numbers.
to get the x variable all by itself, we need to "move" its extra baggage to the other side of the (=) sign.
We first start off by subtracting 6 from both side.
6-6 -3x = 0-6
0 - 3x = -6
-3x = -6
does this make sense so far?

- anonymous

I get so confused when it is a negative I am writing it down step by step

- anonymous

I got it no negative number

- amistre64

good :)
now we are left with -3x = -6 and in order to get rid of the "-3" that is hugging up on that x, we need to remove it by "division" like this.
(-3/-3)x = -6/-3
because any number divided by itself equals 1 we get:
(1)x = -6/-3
x = -6/-3
we should also know by now that whenever we have 2 negatives they become a positive. to give us:
x = 6/3 = 2
So we now that when x = 2 we get a value of zero under the radical:
sqrt(6 -3(2)) = sqrt(6-6) = sqrt(0)
anything to the left of this value is a negative and bad for our problem. Anything that is to the right of this value is a positive and is good with our problem.
<..............2..............>
------- 0 ++++++
So our domain is all the values for x such that x is equal to or greater than 2. or in notation:
[2,inf)

- anonymous

I got it that is what i got to
if i have a problem \[\sqrt{x-1} divided by 1\]

- amistre64

this here is a fraction with a "1" in its bottom part, a 1 in the denominator correct?

- anonymous

1 is the numerator

- amistre64

sqrt(x-1)
-------- is what you have written; is this wrong?
1

- anonymous

yes the one is on top

- anonymous

I got [1,inf)

- amistre64

is there a divided by sign between the square root and that "1" ?
sqrt(x-1) [divide sign] 1 ???

- amistre64

[1,inf) is correct :)

- anonymous

\[1 \over \sqrt{x-1?}\]

- amistre64

ahhhh, then in this case, we have another concern to worry about, the value of the bottom cannot equal zero.

- anonymous

okay why

- amistre64

so we have to restrict our domain of x to all values other than zero, and becasue we have a square root there, we also have to limit it to all values that are positive.

- amistre64

the expression 1/0 has no value in math. it is meaningless. and cannot be used to produce anything with meaning.
5 + 1/0 = ???? cant be solved since 1/0 has no value.

- anonymous

so it is (-inf,inf)

- amistre64

we still have the radical sign to worry about also, which means that we cannot have a value that is less than 0.
All in all, the value of x in: sqrt(x-1) cannot make the inside go negative; and because it is all in the denominator, that value of: sqrt(x-1) cannot be equal to zero.
So we get this:
sqrt(x-1) has the domain [1,inf)
but since x=1 makes this problem = zero: we need to remove it from the domain as well.
so we get:
(1,inf)

- anonymous

on another problem they have 1 over\[\sqrt{2x+2}\] and came up with -1/2,inf

- anonymous

is that because it was 2x+1

- amistre64

1
---------- correct?
sqrt(2x + 2)

- anonymous

never mind I needed to read on

- anonymous

Thank you you have been a lot of help very smart person

- amistre64

thanx :) and youre welcome

- anonymous

I think I can pass this part of my finals

- amistre64

yay!!!

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