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mathcalculus Group Title

where is everyone?!? help please! =( find the extrema: 4x^3-13x^2+4x+1 max and min: [1,4] [0,2]

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    so i found the derivative: 12x^2 -26x +4

    • one year ago
  2. mathcalculus Group Title
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    correct?

    • one year ago
  3. heradog Group Title
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    I could maybe figure it out for myself on my calculator but I don't know how that would help, are you in webworks?

    • one year ago
  4. mathcalculus Group Title
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    yup.

    • one year ago
  5. mathcalculus Group Title
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    do you know hot to factor: 12x^2 -26x +4

    • one year ago
  6. heradog Group Title
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    I'll try and see if I can help but can't promise so yeah you have the derivative right

    • one year ago
  7. mathcalculus Group Title
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    not on calculator though... i just need to know just in case.

    • one year ago
  8. mathcalculus Group Title
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    i must factor first in order to = to 0

    • one year ago
  9. heradog Group Title
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    so to find max and min you find the derivative set it = to 0 and solve?

    • one year ago
  10. mathcalculus Group Title
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    yes i have to find the critical points.. then evaluate the point to original function

    • one year ago
  11. mathcalculus Group Title
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    then evaluate with the endpoints. then i'll find max and min

    • one year ago
  12. heradog Group Title
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    so you have 1/6 and 2 for the critical points right?

    • one year ago
  13. mathcalculus Group Title
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    @campbell_st @precal @heradog the more help the better! :( i really need to figure this problem out.

    • one year ago
  14. mathcalculus Group Title
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    well.... i don't know how to factor this exactly: 12x^2 -26x +4

    • one year ago
  15. mathcalculus Group Title
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    since derivative is right... then how do i factor it out?

    • one year ago
  16. mathcalculus Group Title
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    @dan815

    • one year ago
  17. precal Group Title
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    |dw:1363327156303:dw|

    • one year ago
  18. mathcalculus Group Title
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    yes @precal that one is right

    • one year ago
  19. precal Group Title
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    |dw:1363327198254:dw|

    • one year ago
  20. heradog Group Title
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    ok so to factor you pull out the recurring elements I so suck at factoring but lets try it you can multiply every element by 2 giving you 2(6x^2-13x+2)

    • one year ago
  21. mathcalculus Group Title
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    hm... i did that also... but then i have to find x . critical points.

    • one year ago
  22. precal Group Title
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    |dw:1363327232310:dw|

    • one year ago
  23. precal Group Title
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    |dw:1363327276945:dw|

    • one year ago
  24. precal Group Title
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    are you given those intervals? [1,4] and [0,2]

    • one year ago
  25. campbell_st Group Title
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    the stationary points are at (x - 2)(12x -2) = 0 so the stationary points are x = 2, x = 1/6 so find the the 2nd derivative and test them for max and min then check you boundary values... an easy way it to graph the curve

    • one year ago
  26. precal Group Title
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    why not just do a sign analysis with the first d/dx, second d/dx states concave up or concave down doesn't 1st d/dx show max and min

    • one year ago
  27. mathcalculus Group Title
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    kindly, can we stick with these steps.... {Differentiate} {Set f '(x) = 0} {Factor} {solve to get the critical numbers}

    • one year ago
  28. mathcalculus Group Title
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    im already used to them...

    • one year ago
  29. precal Group Title
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    |dw:1363327423329:dw|

    • one year ago
  30. mathcalculus Group Title
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    @precal what did you do after 2(6x^2-13x+2) ? i dont understand how you got the x^2....etc after...

    • one year ago
  31. precal Group Title
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    |dw:1363327470127:dw|

    • one year ago
  32. precal Group Title
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    |dw:1363327620907:dw|it is an old jedi trick called slide and divide, makes factoring easier to do

    • one year ago
  33. campbell_st Group Title
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    the attached image may help

    • one year ago
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  34. precal Group Title
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    max at x=1/6 and min at x=2

    • one year ago
  35. mathcalculus Group Title
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    |dw:1363327660835:dw|

    • one year ago
  36. precal Group Title
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    |dw:1363327725416:dw|

    • one year ago
  37. mathcalculus Group Title
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    what i meant t say how? i know you said a trick way but i don't know how... :/

    • one year ago
  38. campbell_st Group Title
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    the 2nd derivative test is used for determining if the point is a max or min... easier than 1st der and change of sign...

    • one year ago
  39. precal Group Title
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    true but if this is for a calculus ab class, they are taught to do 1st d/dx to determine max and min

    • one year ago
  40. mathcalculus Group Title
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    stilll dont understand .. :(

    • one year ago
  41. precal Group Title
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    just factor using whatever technique or strategy you were taught, factoring is a basic skill you will need to for future math courses. I just used a technique I like that helps me factor faster......

    • one year ago
  42. campbell_st Group Title
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    ok... here is how I factor it and it works every time given \[ax^2 + bx + c = 0\] multiply a by c in your question 6 times 2 = 12 find the factors of 12 that add to -13....-12, -1 then you have (ax + factor1)(ax + factor 2) (6x -12)(6x -1) -------------------------- -------------- = 0 a 6 find the common factors in the numerator 6(x -2)(6x -1) = 0 ----------------- 6 cancel the common factors leaves (x -2)(6x -1) = 0 easy to solve

    • one year ago
  43. precal Group Title
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    like I said earlier, use whatever works for you. This problem is beyond factoring......

    • one year ago
  44. precal Group Title
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    kindly, can we stick with these steps.... {Differentiate} {Set f '(x) = 0} {Factor} {solve to get the critical numbers} btw I did utilize the steps above

    • one year ago
  45. mathcalculus Group Title
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    yes i used that too. i mea im fine with everything except factoring now.

    • one year ago
  46. mathcalculus Group Title
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    i know it was a basic thing we have to know but i forgot. so I'm trying to find an easy way to learn.

    • one year ago
  47. precal Group Title
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    well I like to use slide and divide

    • one year ago
  48. mathcalculus Group Title
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    i never used that technique before

    • one year ago
  49. precal Group Title
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    http://mrsgalgebra.pbworks.com/w/page/12019090/Slide%20and%20Divide%20Method%20for%20Factoring%20Polynomials

    • one year ago
  50. precal Group Title
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    look it up, you will find many videos on it on youtube. good luck. factoring gets easier over time.

    • one year ago
  51. mathcalculus Group Title
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    thanks :)

    • one year ago
  52. precal Group Title
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    yw

    • one year ago
  53. mathcalculus Group Title
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    @precal

    • one year ago
  54. mathcalculus Group Title
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    i put the answer: [1,4] max=65 min= -11 and for [0,2] max= 196 min= -11 what's wrong with the answer?

    • one year ago
  55. mathcalculus Group Title
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    @dan815

    • one year ago
  56. mathcalculus Group Title
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    somebody help pls!

    • one year ago
  57. precal Group Title
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    are you looking for absolute max or absolute min? are you looking for global or local max or min? seems like you are doing something else other than looking for critical numbers

    • one year ago
  58. precal Group Title
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    gotta go will help you tomorrow, it is very late for me at the moment (past 2 am)

    • one year ago
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