## anonymous one year ago Find the domain of the function. g(x)=^4sq.rt x^2+2x

1. anonymous

yep that's what it's supposed to be

2. IrishBoy123

scan or link the question pls

3. IrishBoy123

sorry, been here done that! just scan or link it :p

4. anonymous

the one you typed is the correct one

5. IrishBoy123

6. anonymous

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?

7. anonymous

$\sqrt[4]{x^2+2x}$

8. anonymous

?

9. freckles

what happens if you have this$\sqrt[4]{-1}$? does this exist over the real numbers?

10. anonymous

I don't think so since you shouldnt have -'s under a radical

11. freckles

great so you want x^2+2x to be positive or zero right?

12. anonymous

yeah

13. freckles

for a number to be positive or equal it must be greater than or equal to 0

14. freckles

so you basically need to solve x^2+2x>=0

15. freckles

to find the domain

16. anonymous

do we just get rid of the radical then?

17. freckles

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

18. freckles

the thing under the radical was x^2+2x

19. freckles

so you want x^2+2x>=0

20. anonymous

it has to be in interval notation

21. anonymous

22. freckles

you would solve x^2+2x>=0

23. freckles

then we can put an interval notation

24. anonymous

x^2>or= -2x

25. anonymous

and then would we square both sides?? because then there'd be a negative on the right

26. freckles

$x^2+2x \ge 0 \\ x(x+2) \ge 0$ draw a number line x(x+2) is zero when x=0 or x+2=0 |dw:1441829182423:dw| test the intervals

27. freckles

|dw:1441829210139:dw|

28. anonymous

OHH so x can equal 0 or -2?

29. anonymous

can;t*

30. freckles

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

31. anonymous

3,-1, and 3?

32. freckles

|dw:1441829448997:dw|

33. freckles

so this tells us x^2+2x is positive on the first and last intervals

34. freckles

in this number line

35. freckles

|dw:1441829505021:dw|

36. anonymous

(-infinity,-2)U(-2,0)U(-2,0)U(0,infinity)?

37. freckles

you need to include both -2 and 0... also why are you include (-2,0) x^2+2x is negative on (-2,0)

38. freckles

remember we were trying to find x such that x^2+2x was positive or zero

39. freckles

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

40. freckles

you should not include (-2,0) do you understand why? you should include -2 and 0 in your solution do you understand this also?

41. anonymous

no I don't :-(

42. freckles

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?

43. freckles

|dw:1441829916519:dw|

44. freckles

|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.

45. freckles

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

46. freckles

|dw:1441830041033:dw|

47. anonymous

oh, so we dont include -2,0 because it would result in a negative

48. freckles

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

49. anonymous

okay so we include those but not (-2,0)

50. freckles

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

51. anonymous

so how would I write that in interval notationn without including (-2,0)