## Christos Group Title Graphing derivatives , and derivative's characteristics http://screencast.com/t/GxfCh0DY So basically this exercise is 100% to be in our final exam today at Calculus , as stated by our teacher so I am trying to master it one year ago one year ago

1. reemii Group Title

and how's it going until now?

2. Christos Group Title

For example On (a) do I have to find roots for only f(x) or f'(x) and f''(x) ?

3. Christos Group Title

it says find roots of f

4. reemii Group Title

the roots of f: the x's such that f(x)=0 they are (graphically) the x's where the graph crosses the x-axis

5. Christos Group Title

ooh first = 3/2

6. Christos Group Title

second,third = (-6 +- sqrt(72))/2

7. Christos Group Title

Which brings me to another problem of mine on how to actually point this number on the graph

8. reemii Group Title

not sure. in the form $$(x+3)^2$$ you should not compute the square, actually they already made the factorization for you. you can just read the answer: second and third roots are equal: x=-3.

9. reemii Group Title

(it's (-6 ± sqrt(0)) / 2 )

10. Christos Group Title

ah yea that's true

11. Christos Group Title

so we have one root for 3/2 and one for -3

12. Christos Group Title

for b I just find f(0) of the function ?

13. reemii Group Title

yes

14. Christos Group Title

-27

15. reemii Group Title

yes again ;)

16. Christos Group Title

So the intervals (-inf,-9) increase (-8,0) decrease (0,inf) increase At least thats what I got

17. reemii Group Title

the derivative of the function is $$6x(x+3)$$. the intervals are (-inf,-3), (-3,0), (0,inf)

18. Christos Group Title

arent we using multiplication rule for the derivative

19. reemii Group Title

yes. whether the values you found are right or wrong, at least verify that the intervals don't leave gaps. (your intervals left a gap between -9 and -8)

20. reemii Group Title

or was it a typo?

21. Christos Group Title

it wasnt a typo but I guess you are right on that one

22. reemii Group Title

directly using the mult. rule: $$((2x-3)(x+3)^2)' = 2(x+3)^2 + (2x-3)2(x+3) = 2(x+3) \times [(x+3)+(2x-3)]$$.

23. reemii Group Title

oops. last one is $2(x+3)\times [(x+3) + (2x-3)] = 2(x+3)[3x] = 6x(x+3)$

24. Christos Group Title

hm let me redo it

25. Christos Group Title

2(x+3)^2 +4x^2 + 12x - 4x -12 up until now all correct?

26. reemii Group Title

i don't think so. do you apply the rule on $$(2x-3)(x^2+6x+9)$$ ? what's your starting point?

27. Christos Group Title

Yes I apply the rule however i start with the product rule first before applying it

28. reemii Group Title

show the first steps plz

29. Christos Group Title

2(x+3)^2+2(x+3(2x-3)

30. Christos Group Title

2(x^2+6x+9)+2(x+3)(2x-3)

31. Christos Group Title

those are correct?

32. reemii Group Title

yes, you made a mistake when computing the product $$2(x+3)(2x-3)$$. this is equal to $$4x^2+6x-18$$.

33. Christos Group Title

Why mistake what do you multiply first with what

34. Christos Group Title

ok ok I got it now for concave up/down (-inf,-1) down (-1,inf) up

35. Christos Group Title

infection points x=-1

36. Christos Group Title

and I am stuck at (f)

37. reemii Group Title

your derivation is correctly done. then it's only multiplication: $2(x^2+6x+9) + 2(x+3)(2x-3)\\ \quad = 2x^2+12x+18 + 2(2x^2-3x+6x-9) \\ \quad = 2x^2+12x+18 + 4x^2 + 6x - 18\\ \quad = 6x^2 + 18x \\ \quad = 6x(x+3)$-> roots are -3 and 0. -> (-inf, -3), (-3,0), (0,inf) are the intervals.

38. Christos Group Title

Yea I see now are my next moves correctly done??

39. reemii Group Title

you didn't obtain the correct expression for f' (you got the wrong roots -> wrong intervals). the inflexion point is at -3/2.

40. Christos Group Title

you mean for f''(x) I got it wrong?

41. Christos Group Title

I damn you are right

42. reemii Group Title

$$f'$$ is unfortunately wrong, so you could not get the right $$f''$$.

43. Christos Group Title

(-inf, -3/2) down (-3/2, inf) up I get it now

44. Christos Group Title

right?

45. reemii Group Title

ouch. no too fast. $$f'(x) = 6x(x+3)$$ with roots -3 and 0. -> (-inf,-3) up, (-3,0) down, (0,+inf) up.

46. reemii Group Title

$$f''(x) = (6(x^2+3x))' = 6(2x+3)$$ with root x=-3/2. -> inflexion point at -3/2.

47. Christos Group Title

Bro I mean concave up down :D I moved a step

48. Christos Group Title

I understood the previews one

49. Christos Group Title

I am just stuck at (f) could you help me out surpass it?

50. reemii Group Title

oh ok. (about concave up down) relative extrema means you have to study what happens at each x such that f'(x)=0, at -3 and at 0 in our case.

51. Christos Group Title

only for f'(x) and f''(x) ?

52. reemii Group Title

the relative extrema (local/global min or max) are only possibly located at the roots of $$f'(x)$$. Look at $$x=-3$$ and say if it's a min, max, or none of these (inflexion point). Do the same with $$x=0$$. You can use $$f''$$ to tell you about the shape of the curve at that particular point.

53. Christos Group Title

is not a min nor a max at -3 0

54. reemii Group Title

you mean (-3,0)? it is a local maximum. Reason: $$f''(-3)<0$$. (or just $$f$$ is increasing before -3 and decreasing after -3.).

55. Christos Group Title

bro you told me to check at f'(x) so which one do I check? do I check both f'(x) and f''(x) ? Do I only use the roots from the first derivative?

56. reemii Group Title

1) to answer the question you have to do some job at each of the points that solve $$f'(x)=0$$. We know that these points are -3 and 0. 2.1) what happens at -3 ? 2.2) what happens at 0 ? To answer 2.1 and 2.2, you can use any means you want. For example, a quick way is to use $$f''$$. $$f''(-3)>0$$ means it's a MIN. $$f''(-3)<0$$ means it's a MAX. $$f''(-3)=0$$ means it's an inflexion point.

57. Christos Group Title

so I dont use -3/2 at all? I can only just use the roots of the first derivative for the second?

58. Christos Group Title

@reemii

59. reemii Group Title

What we know is that -3/2 is an inflexion point. But it happens (often?) that it is not an extrema. Drawing: |dw:1369665423749:dw|

60. reemii Group Title

so you don't need to look at it at all in this question (f). You would have to look at it if it was solution of $$f'(x)=0$$. But it is NOT.

61. Christos Group Title

Ok and the last and hardest part for me can you help me analise/make the graph?? Would it be possible??

62. reemii Group Title

$$x=-3$$ -> $$f''(-3) = 6(2*(-3)+3) < 0$$ -> local max $$x=0$$ -> $$f''(0) = 6(0 + 3)>0$$ -> local min. Ok

63. reemii Group Title

1) Always start with : put points of $$f$$ that are easy: the roots, the intercept. 2) draw the missing part using the info in $$f'$$ and $$f''$$.

64. Christos Group Title

Kinda can you show me what you do to graph it? I have like 5 minutes before the exams that's the most I can get atm

65. reemii Group Title

first step: a few points |dw:1369666161790:dw|

66. reemii Group Title

then join using use $$f''$$ at the extrema (-3 and 0) |dw:1369666278751:dw|

67. reemii Group Title

join in a healthy way (without forgetting hte inflexion point) |dw:1369666362664:dw|

68. reemii Group Title

use that order: 1)fixed points, 2)concavity at extrema 3) join. (pay attention to inflexion point, but that is less important than the rest)