Find the domain of the function. g(x)=^4sq.rt x^2+2x

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Find the domain of the function. g(x)=^4sq.rt x^2+2x

Mathematics
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yep that's what it's supposed to be
scan or link the question pls
sorry, been here done that! just scan or link it :p

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the one you typed is the correct one
scan or link your question no point is answering to ghosts
i don't know how to scan it and the link wont work since it's under my account name.. you already know what the problem looks like, you sent it yourself, so it should be good enough?
\[\sqrt[4]{x^2+2x}\]
?
what happens if you have this\[\sqrt[4]{-1}\]? does this exist over the real numbers?
I don't think so since you shouldnt have -'s under a radical
great so you want x^2+2x to be positive or zero right?
yeah
for a number to be positive or equal it must be greater than or equal to 0
so you basically need to solve x^2+2x>=0
to find the domain
do we just get rid of the radical then?
you told me just a sec ago that we didn't want the thing under the radical to be negative which means we want it to be zero or positive
the thing under the radical was x^2+2x
so you want x^2+2x>=0
it has to be in interval notation
so would I start with (0,_)
you would solve x^2+2x>=0
then we can put an interval notation
x^2>or= -2x
and then would we square both sides?? because then there'd be a negative on the right
\[x^2+2x \ge 0 \\ x(x+2) \ge 0\] draw a number line x(x+2) is zero when x=0 or x+2=0 |dw:1441829182423:dw| test the intervals
|dw:1441829210139:dw|
OHH so x can equal 0 or -2?
can;t*
let me know when you have tested the intervals and yes x=-2 and x=0 will be included in the solution for the domain but we have to also consider the numbers around them
3,-1, and 3?
|dw:1441829448997:dw|
so this tells us x^2+2x is positive on the first and last intervals
in this number line
|dw:1441829505021:dw|
(-infinity,-2)U(-2,0)U(-2,0)U(0,infinity)?
you need to include both -2 and 0... also why are you include (-2,0) x^2+2x is negative on (-2,0)
remember we were trying to find x such that x^2+2x was positive or zero
we found x^2+2x was zero at x=-2, x=0 we found x^2+2x was positive for values to the left of -2, also values to right of 0
you should not include (-2,0) do you understand why? you should include -2 and 0 in your solution do you understand this also?
no I don't :-(
hmm... so do you remember when we tested the intervals earlier and for the member that represented the set of numbers on (-2,0) we got a negative number?
|dw:1441829916519:dw|
|dw:1441829938691:dw| I put check marks on the intervals we want because we were looking for intervals where x^2+2x was positive I put an X mark on intervals we don't want because we were not looking for when x^2+2x was negative.
the 4 root of 0 exists and it is 0 so x^2+2x=0 will satisfy our domain x(x+2)=0 when x=0 and x=-2 these numbers have to be a part of the domain
|dw:1441830041033:dw|
oh, so we dont include -2,0 because it would result in a negative
no we don't include (-2,0) because any number from that set would result in a negative we do include -2 and 0 though because pluggin in those numbers result in zero
okay so we include those but not (-2,0)
yes any number from (-2,0) when pluggin in to x^2+2x will give you a negative number you have already said the 4th root of a negative number doesn't exist over the real numbers
so how would I write that in interval notationn without including (-2,0)

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