1. mikoza

is (-infinity ,-4]U[8,+infinity)

2. Dodo1

i entered it and it told me that it was wrong

3. davisla

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 $\sqrt{(x-8)(x+4}$ the radicand is (x-8)(x+4) set up the inequality $(x-8)(x+4)\ge0$

4. davisla

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?

5. mikoza

dear davisla (x-4)(x+8)is not equal to (x-8)(x+4)

6. davisla

you are correct ... so for what values of x is (x - 8)(x + 4) = 0?

7. Dodo1

I thought that X=-8. and X=+4. I am not sure how the infinity blacket works

8. Dodo1

sorry I mean X=+8 and X=-4

9. davisla

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.

10. Dodo1

so the second interval should be (8,infinity)?

11. davisla

the second is (-4, 8) the third is (8, infinity)

12. davisla

in the second interval, I would choose 0 to plug in since it is easy

13. Dodo1

Ok so, the answer should be (-infinity,-4)U(-4,0)U(0,8)U(8,infinity)?

14. davisla

no, when you plug 0 in you get: (0-8)(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?

15. Dodo1

I got positive.

16. davisla

so that interval is in our domain ... now, what about -4 and 8? Can the expression = 0?

17. davisla

we can take the sq root of zero, so we do want to include those endpoints in our intervals

18. davisla

we wind up with the intervals given by someone else in the beginning and that is correct. $(-\infty,-4]\cup[8,\infty)$

19. davisla

I am wondering if it didn't like the way you input ... did you put a + in front of the second infinity?