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Christos

  • one year ago

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

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  1. reemii
    • one year ago
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    and how's it going until now?

  2. Christos
    • one year ago
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    For example On (a) do I have to find roots for only f(x) or f'(x) and f''(x) ?

  3. Christos
    • one year ago
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    it says find roots of f

  4. reemii
    • one year ago
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    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
    • one year ago
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    ooh first = 3/2

  6. Christos
    • one year ago
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    second,third = (-6 +- sqrt(72))/2

  7. Christos
    • one year ago
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    Which brings me to another problem of mine on how to actually point this number on the graph

  8. reemii
    • one year ago
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    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
    • one year ago
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    (it's (-6 ± sqrt(0)) / 2 )

  10. Christos
    • one year ago
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    ah yea that's true

  11. Christos
    • one year ago
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    so we have one root for 3/2 and one for -3

  12. Christos
    • one year ago
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    for b I just find f(0) of the function ?

  13. reemii
    • one year ago
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    yes

  14. Christos
    • one year ago
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    -27

  15. reemii
    • one year ago
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    yes again ;)

  16. Christos
    • one year ago
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    So the intervals (-inf,-9) increase (-8,0) decrease (0,inf) increase At least thats what I got

  17. reemii
    • one year ago
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    the derivative of the function is \(6x(x+3)\). the intervals are (-inf,-3), (-3,0), (0,inf)

  18. Christos
    • one year ago
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    arent we using multiplication rule for the derivative

  19. reemii
    • one year ago
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    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
    • one year ago
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    or was it a typo?

  21. Christos
    • one year ago
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    it wasnt a typo but I guess you are right on that one

  22. reemii
    • one year ago
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    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
    • one year ago
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    oops. last one is \[ 2(x+3)\times [(x+3) + (2x-3)] = 2(x+3)[3x] = 6x(x+3)\]

  24. Christos
    • one year ago
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    hm let me redo it

  25. Christos
    • one year ago
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    2(x+3)^2 +4x^2 + 12x - 4x -12 up until now all correct?

  26. reemii
    • one year ago
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    i don't think so. do you apply the rule on \((2x-3)(x^2+6x+9)\) ? what's your starting point?

  27. Christos
    • one year ago
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    Yes I apply the rule however i start with the product rule first before applying it

  28. reemii
    • one year ago
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    show the first steps plz

  29. Christos
    • one year ago
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    2(x+3)^2+2(x+3(2x-3)

  30. Christos
    • one year ago
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    2(x^2+6x+9)+2(x+3)(2x-3)

  31. Christos
    • one year ago
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    those are correct?

  32. reemii
    • one year ago
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    yes, you made a mistake when computing the product \(2(x+3)(2x-3)\). this is equal to \(4x^2+6x-18\).

  33. Christos
    • one year ago
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    Why mistake what do you multiply first with what

  34. Christos
    • one year ago
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    ok ok I got it now for concave up/down (-inf,-1) down (-1,inf) up

  35. Christos
    • one year ago
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    infection points x=-1

  36. Christos
    • one year ago
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    and I am stuck at (f)

  37. reemii
    • one year ago
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    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
    • one year ago
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    Yea I see now are my next moves correctly done??

  39. reemii
    • one year ago
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    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
    • one year ago
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    you mean for f''(x) I got it wrong?

  41. Christos
    • one year ago
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    I damn you are right

  42. reemii
    • one year ago
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    \(f'\) is unfortunately wrong, so you could not get the right \(f''\).

  43. Christos
    • one year ago
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    (-inf, -3/2) down (-3/2, inf) up I get it now

  44. Christos
    • one year ago
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    right?

  45. reemii
    • one year ago
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    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
    • one year ago
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    \(f''(x) = (6(x^2+3x))' = 6(2x+3)\) with root x=-3/2. -> inflexion point at -3/2.

  47. Christos
    • one year ago
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    Bro I mean concave up down :D I moved a step

  48. Christos
    • one year ago
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    I understood the previews one

  49. Christos
    • one year ago
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    I am just stuck at (f) could you help me out surpass it?

  50. reemii
    • one year ago
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    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
    • one year ago
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    only for f'(x) and f''(x) ?

  52. reemii
    • one year ago
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    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
    • one year ago
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    is not a min nor a max at -3 0

  54. reemii
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    so I dont use -3/2 at all? I can only just use the roots of the first derivative for the second?

  58. Christos
    • one year ago
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    @reemii

  59. reemii
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Ok and the last and hardest part for me can you help me analise/make the graph?? Would it be possible??

  62. reemii
    • one year ago
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    \(x=-3\) -> \(f''(-3) = 6(2*(-3)+3) < 0\) -> local max \(x=0\) -> \(f''(0) = 6(0 + 3)>0\) -> local min. Ok

  63. reemii
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    first step: a few points |dw:1369666161790:dw|

  66. reemii
    • one year ago
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    then join using use \(f''\) at the extrema (-3 and 0) |dw:1369666278751:dw|

  67. reemii
    • one year ago
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    join in a healthy way (without forgetting hte inflexion point) |dw:1369666362664:dw|

  68. reemii
    • one year ago
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    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)

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