## Christos 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 9 months ago 9 months ago

1. reemii

and how's it going until now?

2. Christos

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

3. Christos

it says find roots of f

4. reemii

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

ooh first = 3/2

6. Christos

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

7. Christos

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

8. reemii

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

(it's (-6 ± sqrt(0)) / 2 )

10. Christos

ah yea that's true

11. Christos

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

12. Christos

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

13. reemii

yes

14. Christos

-27

15. reemii

yes again ;)

16. Christos

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

17. reemii

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

18. Christos

arent we using multiplication rule for the derivative

19. reemii

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

or was it a typo?

21. Christos

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

22. reemii

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

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

24. Christos

hm let me redo it

25. Christos

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

26. reemii

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

27. Christos

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

28. reemii

show the first steps plz

29. Christos

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

30. Christos

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

31. Christos

those are correct?

32. reemii

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

33. Christos

Why mistake what do you multiply first with what

34. Christos

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

35. Christos

infection points x=-1

36. Christos

and I am stuck at (f)

37. reemii

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

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

39. reemii

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

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

41. Christos

I damn you are right

42. reemii

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

43. Christos

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

44. Christos

right?

45. reemii

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

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

47. Christos

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

48. Christos

I understood the previews one

49. Christos

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

50. reemii

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

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

52. reemii

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

is not a min nor a max at -3 0

54. reemii

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

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

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

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

58. Christos

@reemii

59. reemii

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

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

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

62. reemii

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

63. reemii

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

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

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

66. reemii

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

67. reemii

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

68. reemii

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)