We need help finding the domain of functions

Question

anonymous · Answer

$$y=3x-2 \over 4x+1$$

mathteacher1729 · Answer

$y = \frac{3x-2}{4x+1}$ is a perfectly valid function EXCEPT whenever the denominator = 0.  

So the question is... what value(s) of x make the denominator = 0? The domain is all real numbers, EXCEPT that value (or those values) of x. :) 

Hope this helps.

anonymous · Answer

$$y = \frac{3x-2}{4x+1}$$

y = \frac{3x-2}{4x+1}

anonymous · Answer

What about , $$y=x ^{2}-5x-6 \over x ^{2}-3x-18$$ how would we find the domain of this?

anonymous · Answer

We don't quite understand how to find the domain

myininaya · Answer

factor the bottom
find when the bottom is zero
(x-6)(x+3)=0
solve for x

anonymous · Answer

so is that the domain? , what about infinites?

myininaya · Answer

when we are looking for domain 
we are looking for some real input that does not have a real output 
is there anything you can enter into your function that would not give you a real output?

anonymous · Answer

we r just so confused! i got x = 6 and x= -3, but that's what x can not equal ?

anonymous · Answer

?

myininaya · Answer

right because then the bottom would be zero

anonymous · Answer

we don't take it to infinity, like (-3,6), (6,$$\infty$$)

myininaya · Answer

the domain is all real numbers except x=-3 and x=6
you can write in interval notation if you like 
(-inf,-3)U(-3,6)U(6,inf)

anonymous · Answer

Thank you so much! so for example $$y=2^{2-x} \over x$$ would the domain just be 0, so (-infin, to infin)?

mathteacher1729 · Answer

afritz -- the domain is all values of x which DO NOT break the function.  

Anytime the denominator is zero the function "breaks". 

So in your most recent example, the domain would be all real numbers EXCEPT zero.

myininaya · Answer

what if you have
$$y=\sqrt{3x-1}$$
do you know how to find domain here?

anonymous · Answer

thank you mathteacher1729, and myininaya, we are not sure, but is it (-infinity, 1/3)U(1/3, infinity)

myininaya · Answer

so
what would happen if I plugged in -3

anonymous · Answer

it would be negative square root -10? , was that first answer wrong or right/ sorry this has always caused me issues.

myininaya · Answer

$$f(-3)=\sqrt{3(-3)-1}=\sqrt{-9-1}=\sqrt{-10}$$
if you have a negative under the square root you do not get a real output
so you want 3x-1 to be positive or neutral
so to find domain you would do 3x-1>=0

myininaya · Answer

$$3x-1 \ge 0$$

we do this because we want 3x-1 to be positive (>0) or neutral (=0)

myininaya · Answer

$$3x \ge 1$$
$$x \ge \frac{1}{3}$$

so in interval notation you would say [1/3,inf)

anonymous · Answer

okay i see, so what about $$y=\sqrt{x-3}-\sqrt{x+3}$$ we have a few more after this, i do apologize!

anonymous · Answer

we also don' t understand how to put it into notation, we have to do it that way for our calculus packet this summer.

anonymous · Answer

like we do not know how to put the  numbers in it.

myininaya · Answer

ok look at the first part of your function
we have $$\sqrt{x-3}$$
what does x-3 have to be in order for you to have a real output?

anonymous · Answer

3? or is that what it can NOT be?

myininaya · Answer

x-3 has to be positive or neutral

x-3> or = 0

$$x-3 \ge 0$$
$$x \ge 3$$

you also have the other part of your function is $$\sqrt{x+3}$$

what does x+3 have to be in order for you to have a real output

anonymous · Answer

$$x \ge -3$$

anonymous · Answer

so what would the interval notation for that be and why?

myininaya · Answer

ok now lets look at this and graph it on a number line we have from the first part 
$$x \ge 3$$ 
and second part we had
$$x \ge -3$$
we don't shade here because the inequalities   don't agree on this area
                       |     |
                      /   /      

                                           *~~~~~~~~~~
-------|--------------|------------
            -3                          3

anonymous · Answer

(-infin, -3]U[3,infin) would this be right?

myininaya · Answer

no the area that is shaded represents the domain
so the domain is [3,inf)

anonymous · Answer

and why is that the only shaded part? sorry to keep asking lol

myininaya · Answer

do you see the inequalites we have up there?
i graphed the inequalities on a number line together

anonymous · Answer

$$y=\sqrt{2x-9} \over 2x+9$$ the bottom would be x $$x \ge -9/2$$ the top would be $$x \ge 9/2$$ so the domain [9/2,infin), and what about the negative , why don't they exist in the notation?

anonymous · Answer

was that correct?

myininaya · Answer

you don't want the bottom to be 0 
so you don't want 2x+9 to be 0
and it is 0 when x=-9/2

you have a square root also you want the inside to be positive or neutral
so you want 2x-9>=0
2x>=9
x>=9/2
                     *~~~~~~~~~~
-----------|-----------
                     9/2

[9/2,inf)
is the domain 
very good! :)

anonymous · Answer

Thank you very much, the only thing i am confused on now is why we do not include [-infint, -9/2) is there a reason we do not include that? 
 and the next equation is 

$$y=x ^{2} + 8x + 12 \over \sqrt[4]{x+5}$$

anonymous · Answer

the numerator of that wold be (x+6)(x+2) so x can not be -6 or -2, but the bottom I have no idea! would it be $$x \ge -5$$ ,

anonymous · Answer

sorry i know that you are busy

anonymous · Answer

I would post that as a new question along with your attempt at an answer.

myininaya · Answer

hey afritz i think you would get more responses if you made a new post every once and awhile like estudier said

anonymous · Answer

ok thanks!  i actually just figured out how to do that!

myininaya · Answer

and right you want x>=-5

myininaya · Answer

and above you wanted to know why we didn't incluse (-inf,-9/2]
what would happen if we entered one of these numbers in the set (or interval)
(-inf,-9/2] into y=sqrt{2x-9}/(2x+9)

anonymous · Answer

so it would be -infinity to -5? correct, and okay i see!

myininaya · Answer

(-5,inf)