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what does y have to be in order for x to exist?

I am not sure? this question is confusing

can you have a negative under the square root?

no

you want 36+4y to be bigger than or equal to zero, correct?

yes correct

so solve this for y: 36+4y>=0 and you will have your range

-9 >= y

how do i write the range

4y>=-36
so y>=-9

what is the domain?

solve for y

to find range we solve for x
to find domain we solve for y
so solve the orignal equation for y

so y >= -9 is the range or domain?

are you still here?

i think you are supposed to solve each of these for y
and think of y as a function of x

but there is something wrong here because you have these each set = 0

that is what my project say

y is range we got y>=-9

now*

not no lol

LOL u guys are great help

second one.
\[4x^2-xy+1=0\]
\[xy=4x^2+1\]
\[y=\frac{4x^2+1}{x}\]
domain all real numbers except 0

so the second one does not have a range

range is also all real numbers except 0 because the numerator is \[4x^2+1\] which cannot be 0

range is [-4,4] you find once you solve for x

oops really? i will try it.

oops o did somehting wrong its (-inf,-4]U[4,inf)

is this for the second problem

the problem that we are now we are finding range

how do you like that. you get \[y^2-16\]under the radical so \[y^2-16\geq 0\]
who knew?

lol satellitle

you remember all that inverse crap right?

so the range is (-inf, -4)U(4,inf)

is the last one \[5x^2y-x+2=0\]?

yes you have the range correct.

ok

if i have the last one correct is is
\[y=\frac{x-2}{5x^2}\]
domain all numbers except 0

and the range?

honestly i am little confused by this. myininaya said solve for x. i can try it.

oh ok!

so i think i will stick to that answer, and go to bed.

you want me to check your range? lol
which one is this?

ok the last one

last one. i have confused myself.

if i didnt make a mistake here is the range

i must be tired if i cannot multiply 4 by 5 by 2!
i best retire.

ok maybe but i am still skeptical. what do you get if x = 2?

hmm you get (2,0)

i should switch back to windows and run maple, graph the thing and then i would be more convinced.

so the range is (-inf,1/40]

i am afraid i do not believe that either.

i believe in it lol

as x goes to infinity
\[\frac{x-2}{5x^2}\] goes to zero

right there is a horizontal asymptote

let x = 1000000

bet you get something closer to 0 than 1/40

oh wait. maybe it is always negative. oops. ok possible

ok i am beginning to believe it. may be always less than 1/40. tomorrow i graph
!

hey satelittle you know what would really convince you?
i do

maple

you know how to find the maximum right?
that derivative crap?

ok here we go.

so you do :) yay!

f'(x)=[5x^2-10x(x-2)]/[(5x^2)]^2=[-5x^2+20x]/(25x^4)
a critical numbers isfour
f(4)=1/40

well it is nice and late and i have had a few beers. but it looks like extreme value at 0 and 4

well 0 doesn't exist in the orignal function

0 is out, 4 gives....
holy carp, what do you know.

lol

learn something new every day. thanks.

who needs maple

me. use it all the time when i am not running linux

well it is a nice friend to have

and so to bed, before i make more of a fool of myself. take care

later

i'm going too gn