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mathcalculus

H-E-L-P Determine the extrema of below on the given interval f(x)=5x^3-61x^2+16x+3 (a) on [0,4] The minimum is and the maximum is . (b) on [-9,9] The minimum is and the maximum is

  • one year ago
  • one year ago

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  1. completeidiot
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    first derivative and set to zero to find the local max and min other points of interest would be the end points

    • one year ago
  2. mathcalculus
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    THE DERIVATIVE IS 15X^2-122X+16

    • one year ago
  3. mathcalculus
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    DO WE FIND ALSO 2ND DERIVATIVE TOO?

    • one year ago
  4. calmat01
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    lol..here we are again...it's back to factoring.

    • one year ago
  5. calmat01
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    You need to find factors of 240, which is the product of 15 and 16 which add to the coefficient of the middle term.

    • one year ago
  6. calmat01
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    If you split the -122x into -120x and -2x, you will get the numbers I was referring to. Then you can factor by grouping.

    • one year ago
  7. mathcalculus
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    yup here we are... where the 240 come from...

    • one year ago
  8. completeidiot
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    i dont believe 2nd derivative is necessary for determining extremas

    • one year ago
  9. calmat01
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    I explained already...240 comes from multiplying the 15(coefficient of squared term) and 16 (the constant).

    • one year ago
  10. mathcalculus
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    ok show it.

    • one year ago
  11. mathcalculus
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    why do you multiply it?

    • one year ago
  12. mathcalculus
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    and the middle term is -122x

    • one year ago
  13. calmat01
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    Because it is not a trinomial with leading coefficient of 1.

    • one year ago
  14. mathcalculus
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    yes. right..

    • one year ago
  15. calmat01
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    That is why I said you want numbers which add to the coeffcient of the middle term.

    • one year ago
  16. calmat01
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    The two numbers which multiply to give you 240 that add to -122 are -120 and -2.

    • one year ago
  17. calmat01
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    So, basic algebra tells us that if you want to factor an equation of the form \[ax ^{2}+bx+c=0 \] we must multiply a*c and find factors of a*c which add to the coeffcient of the middle term, in this case -122.

    • one year ago
  18. calmat01
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    And those factors, as I already mentioned, are -120 and -2. So, just attach the x to each of them, group the first two and last two terms in separate parenthesis and factor by grouping.

    • one year ago
  19. calmat01
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    \[15x ^{2}-122x+16=15x ^{2}-120x-2x+16=0\] Grouping them as I have instructed, you get \[(15x ^{2}-120x) + (-2x+16)=0\]

    • one year ago
  20. calmat01
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    Now just factor a 15x from the first parenthesis, and a -2 from the second parenthesis, and what do you notices when you do that?

    • one year ago
  21. mathcalculus
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    oh i got it... you just separated them ..

    • one year ago
  22. mathcalculus
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    can you separate them by anything? ex: -61x ?

    • one year ago
  23. mathcalculus
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    since 122/2 = 61..

    • one year ago
  24. mathcalculus
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    nvm i understand.. you want something you can easily factor with 15 and 2

    • one year ago
  25. mathcalculus
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    so i get.. x= 8 and 2/15

    • one year ago
  26. calmat01
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    There you go.

    • one year ago
  27. mathcalculus
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    which one is max and min?

    • one year ago
  28. mathcalculus
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    wait the intervals are [0,4] what do we do with it?

    • one year ago
  29. calmat01
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    Check to see what happens to the derivative to the left and right of the point x=2/15 since x=8 is not in the interval.

    • one year ago
  30. mathcalculus
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    O_o

    • one year ago
  31. mathcalculus
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    true.. but what do you mean derivative to the left and right?

    • one year ago
  32. completeidiot
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    plug x=0 and x=4 into the original equation in order to find the extrema points

    • one year ago
  33. calmat01
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    You can tell whether you have a max or min by what happens to the derivative to the left and right of the critical point. If f'(x) > o, your function is increasing, if f'(x)<0, your function is decreasing.

    • one year ago
  34. calmat01
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    |dw:1363299185209:dw|

    • one year ago
  35. calmat01
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    Hope that makes sense.

    • one year ago
  36. mathcalculus
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    f(0)= 3 and f(4) -589

    • one year ago
  37. mathcalculus
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    okay those are the extrema points.. but how about the work for x= 8 and x= 2/15?

    • one year ago
  38. mathcalculus
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    @calmat01

    • one year ago
  39. mathcalculus
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    its wrong....

    • one year ago
  40. mathcalculus
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    i see how you got the 16 and -232... but those aren;t the min or max of the interval.

    • one year ago
  41. mathcalculus
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    i put 16 as min ad -232 as max. its wrong.

    • one year ago
  42. calmat01
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    No, I was telling you that the max occurs at x=2/15, not that the values I used were the max or min. You still have to determine the max by substituting your value for x=2/15.

    • one year ago
  43. calmat01
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    Because the graph at x=2/15 changes direction...it goes from increasing to dercreasing thus identifying a max.

    • one year ago
  44. mathcalculus
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    okay so far i have f'(0)=16 f'(4)=-232 and the x=8 and the other x= 2/15

    • one year ago
  45. dan815
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    take first dervivative and find crit points and plug them into 2nd derivative to see if those crit ppts are maxima(negative for 2nd derv) or minima(positive)

    • one year ago
  46. mathcalculus
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    i would appreciate if i can see some work please.

    • one year ago
  47. dan815
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    do u understand what i said

    • one year ago
  48. mathcalculus
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    yeah but i'm more of a visual person..

    • one year ago
  49. mathcalculus
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    i understand what you're saying though.

    • one year ago
  50. dan815
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    http://www.wolframalpha.com/input/?i=16-122+x%2B15+x%5E2&lk=1&a=ClashPrefs_*Math-

    • one year ago
  51. dan815
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    that is the graph of ur derivative

    • one year ago
  52. dan815
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    to be honest if ur visual u should realize its a 3rd degree func to 2nd degree so its like this|dw:1363301376615:dw|

    • one year ago
  53. dan815
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    so say you find f''(x) = 0 then you will know anything grreater than that x value will give u minimas and anything less than 0 for 2nd derv will give you maximas because they are inflection points

    • one year ago
  54. dan815
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    these are inflection points say for your base function x^3 ...+...+...=y

    • one year ago
  55. dan815
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    |dw:1363301676110:dw|

    • one year ago
  56. dan815
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    so whats its basically asking you is to find the critical pts and tell you if they are concave up or down so for the example above u know the critical pts are these|dw:1363301848503:dw|

    • one year ago
  57. dan815
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    where your slope is 0 right so those are crit pts but do those crit points lie where the graph is concave up or down

    • one year ago
  58. dan815
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    for the first crit pt you can see its on the part of the graph where it is concave down and the 2nd one is on the part of the graph where its concave up

    • one year ago
  59. dan815
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    make sense now?

    • one year ago
  60. dan815
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    but lets suppose this is your interval |dw:1363302163817:dw|

    • one year ago
  61. dan815
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    so for the first critical point you know thats a maxima between the interval where your interval is x between those 2 vertical lines, but your minimum in that interval wont be the 2nd critival point because that is outside of the interval rather you will have to check the ends of the interval to see which one is of less value and that is your minumum

    • one year ago
  62. mathcalculus
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    oh okay thank you=]

    • one year ago
  63. mathcalculus
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    i see what method you used.

    • one year ago
  64. mathcalculus
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    so is the minim here: -589 and the max -1213??

    • one year ago
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