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mathcalculus

  • 2 years ago

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

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  1. mathcalculus
    • 2 years ago
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    so i found the derivative: 12x^2 -26x +4

  2. mathcalculus
    • 2 years ago
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    correct?

  3. heradog
    • 2 years ago
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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?

  4. mathcalculus
    • 2 years ago
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    yup.

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

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

  7. mathcalculus
    • 2 years ago
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    not on calculator though... i just need to know just in case.

  8. mathcalculus
    • 2 years ago
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    i must factor first in order to = to 0

  9. heradog
    • 2 years ago
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    so to find max and min you find the derivative set it = to 0 and solve?

  10. mathcalculus
    • 2 years ago
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    yes i have to find the critical points.. then evaluate the point to original function

  11. mathcalculus
    • 2 years ago
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    then evaluate with the endpoints. then i'll find max and min

  12. heradog
    • 2 years ago
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    so you have 1/6 and 2 for the critical points right?

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

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

  15. mathcalculus
    • 2 years ago
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    since derivative is right... then how do i factor it out?

  16. mathcalculus
    • 2 years ago
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    @dan815

  17. precal
    • 2 years ago
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    |dw:1363327156303:dw|

  18. mathcalculus
    • 2 years ago
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    yes @precal that one is right

  19. precal
    • 2 years ago
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    |dw:1363327198254:dw|

  20. heradog
    • 2 years ago
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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)

  21. mathcalculus
    • 2 years ago
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    hm... i did that also... but then i have to find x . critical points.

  22. precal
    • 2 years ago
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    |dw:1363327232310:dw|

  23. precal
    • 2 years ago
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    |dw:1363327276945:dw|

  24. precal
    • 2 years ago
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    are you given those intervals? [1,4] and [0,2]

  25. campbell_st
    • 2 years ago
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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

  26. precal
    • 2 years ago
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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

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

  28. mathcalculus
    • 2 years ago
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    im already used to them...

  29. precal
    • 2 years ago
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    |dw:1363327423329:dw|

  30. mathcalculus
    • 2 years ago
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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...

  31. precal
    • 2 years ago
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    |dw:1363327470127:dw|

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

  33. campbell_st
    • 2 years ago
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    the attached image may help

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  34. precal
    • 2 years ago
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    max at x=1/6 and min at x=2

  35. mathcalculus
    • 2 years ago
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    |dw:1363327660835:dw|

  36. precal
    • 2 years ago
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    |dw:1363327725416:dw|

  37. mathcalculus
    • 2 years ago
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    what i meant t say how? i know you said a trick way but i don't know how... :/

  38. campbell_st
    • 2 years ago
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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...

  39. precal
    • 2 years ago
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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

  40. mathcalculus
    • 2 years ago
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    stilll dont understand .. :(

  41. precal
    • 2 years ago
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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......

  42. campbell_st
    • 2 years ago
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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

  43. precal
    • 2 years ago
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    like I said earlier, use whatever works for you. This problem is beyond factoring......

  44. precal
    • 2 years ago
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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

  45. mathcalculus
    • 2 years ago
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    yes i used that too. i mea im fine with everything except factoring now.

  46. mathcalculus
    • 2 years ago
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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.

  47. precal
    • 2 years ago
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    well I like to use slide and divide

  48. mathcalculus
    • 2 years ago
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    i never used that technique before

  49. precal
    • 2 years ago
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    look it up, you will find many videos on it on youtube. good luck. factoring gets easier over time.

  50. mathcalculus
    • 2 years ago
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    thanks :)

  51. precal
    • 2 years ago
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    yw

  52. mathcalculus
    • 2 years ago
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    @precal

  53. mathcalculus
    • 2 years ago
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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?

  54. mathcalculus
    • 2 years ago
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    @dan815

  55. mathcalculus
    • 2 years ago
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    somebody help pls!

  56. precal
    • 2 years ago
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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

  57. precal
    • 2 years ago
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    gotta go will help you tomorrow, it is very late for me at the moment (past 2 am)

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