## anonymous one year ago Use synthetic division to determine whether the number k is an upper or lower bound (as specified) for the real zeros of the function f. k = 4; f(x) = 2x^3 - 2x^2 - 3x - 5; Lower bound?

1. anonymous

Yes or no are the answer choices

2. anonymous

@Loser66

3. Loser66

So, you have to solve for the zeros of the function. What are they?

4. anonymous

I'm honestly not sure how to find them

5. Loser66

use demos graph to see where they cut x-axis

6. anonymous

(2.2,0)

7. anonymous

So would the answer be no since it's (2.2,0) and 4 would be a higher bound and not lower bound?

8. Loser66

we don't have 0. please check

9. anonymous

what do you mean? it cut the x-axis at 2.2

10. Loser66

but you can use it. now find f(0) =?

11. Loser66

replace 0 into f(x) , what do you get?

12. Loser66

|dw:1442536923437:dw|

13. anonymous

-5

14. Loser66

ok, so f(0) =-5 < 0 , right? now f (2) =?

15. anonymous

-3

16. anonymous

So what does that mean

17. Loser66

that mean from 0 to 2 you still not pass the x-axis, since f(x) <0 for both. Now test f(3)

18. anonymous

22

19. Loser66

yes, so from 2 to 3 , f(x) go from negative to positive, that is you have one zero on that interval, right?

20. Loser66

Now, use synthetic,

21. Loser66

forget it, short cut

22. anonymous

okay haha

23. Loser66

|dw:1442537533408:dw|

24. Loser66

now, fill up

25. anonymous

Again, the lesson didnt really explain this stuff so I'm not sure how to use that

26. Loser66

you don't know synthetic?

27. anonymous

not completely, looking it up now trying to learn real quick

28. Loser66

|dw:1442537680182:dw|

29. Loser66

|dw:1442537714109:dw|

30. Loser66

now 4 *6 =?

31. anonymous

24

32. Loser66

|dw:1442537779766:dw| got it?

33. Loser66

the last one, you do, the same

34. anonymous

would it be 4*21 = 84 or 4*24 =96?

35. Loser66

|dw:1442537857152:dw|

36. Loser66

you take 4 * the bottom row, put the result on the middle row of the new column, then add them up to get the new number.

37. anonymous

oh okay i see

38. anonymous

that's easy then haha

39. Loser66

|dw:1442538058710:dw|

40. Loser66

ok, now, 4 >0 , right?

41. anonymous

yes

42. Loser66

look at the bottom row, they are 2,6,21,79 right?

43. anonymous

yes

44. Loser66

and all them are positive, right?

45. anonymous

yes

46. Loser66

Hence 4 is upper bound for the zeros

47. anonymous

woah i see, a lot goes into it

48. anonymous

so back to the original question the answer would be no

49. Loser66

yup

50. anonymous

ok thanks for all of the help, what would happen if one of them was negative?

51. Loser66

you test, if k <0, the bottom line is +, -,+,- alternatively, then k is lower bound

52. Loser66

learn from it http://www.howe-two.com/mathematicat/bound.html