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anonymous
 4 years ago
Please help me.
Domain of sqrt ((x8)(x+4))
thanks
anonymous
 4 years ago
Please help me. Domain of sqrt ((x8)(x+4)) thanks

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is (infinity ,4]U[8,+infinity)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i entered it and it told me that it was wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Answers 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 \[(x8)(x+4)\ge0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now, 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?

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry I mean X=+8 and X=4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0once 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the second interval should be (8,infinity)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the second is (4, 8) the third is (8, infinity)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in the second interval, I would choose 0 to plug in since it is easy

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no, 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?

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

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

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

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