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mikozaBest ResponseYou've already chosen the best response.0
is (infinity ,4]U[8,+infinity)
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
i entered it and it told me that it was wrong
 one year ago

davislaBest ResponseYou've already chosen the best response.0
Answers without solutions don't help understanding so then you can't do the next one or the one on your exam. The value we are taking the square root of must be greater than or equal to 0, so in the expression \[\sqrt{(x8)(x+4}\] the radicand is (x8)(x+4) set up the inequality \[(x8)(x+4)\ge0\]
 one year ago

davislaBest ResponseYou've already chosen the best response.0
Now, the only way we can go from positive to negative is to pass through 0 and the only way we can go from negative to positive is to pass through 0, so find the zeros. For what values of x is (x  4)(x+8) = 0?
 one year ago

mikozaBest ResponseYou've already chosen the best response.0
dear davisla (x4)(x+8)is not equal to (x8)(x+4)
 one year ago

davislaBest ResponseYou've already chosen the best response.0
you are correct ... so for what values of x is (x  8)(x + 4) = 0?
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
I thought that X=8. and X=+4. I am not sure how the infinity blacket works
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
sorry I mean X=+8 and X=4
 one year ago

davislaBest ResponseYou've already chosen the best response.0
once you find the zeros, you know where it COULD change signs. That does not mean it does. So, from neg infinity to 4 it NEVER changes signs, from 4 to 8 it NEVER changes signs and from 8 to infinity it NEVER changes signs Plug in a number in each of those 3 intervals to determine the sign of the expression in that interval. Try 5: (5  4)(5 + 4) = 9 (1) which is +9 ... but all we care about is that it is positive. So in the interval (infinity, 4) it is positive  that is okay so that is part of the domain. Now try a number in the second interval and see if it is positive or negative.
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
so the second interval should be (8,infinity)?
 one year ago

davislaBest ResponseYou've already chosen the best response.0
the second is (4, 8) the third is (8, infinity)
 one year ago

davislaBest ResponseYou've already chosen the best response.0
in the second interval, I would choose 0 to plug in since it is easy
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
Ok so, the answer should be (infinity,4)U(4,0)U(0,8)U(8,infinity)?
 one year ago

davislaBest ResponseYou've already chosen the best response.0
no, when you plug 0 in you get: (08)(0+4) = 8(4) and that is a negative number  so that interval is no good and is not in your domain what about the last interval? do you get a positive or a negative?
 one year ago

davislaBest ResponseYou've already chosen the best response.0
so that interval is in our domain ... now, what about 4 and 8? Can the expression = 0?
 one year ago

davislaBest ResponseYou've already chosen the best response.0
we can take the sq root of zero, so we do want to include those endpoints in our intervals
 one year ago

davislaBest ResponseYou've already chosen the best response.0
we wind up with the intervals given by someone else in the beginning and that is correct. \[(\infty,4]\cup[8,\infty)\]
 one year ago

davislaBest ResponseYou've already chosen the best response.0
I am wondering if it didn't like the way you input ... did you put a + in front of the second infinity?
 one year ago
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