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mathcalculus

  • one year ago

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

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  1. completeidiot
    • one year ago
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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

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

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

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

  5. calmat01
    • one year ago
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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.

  6. calmat01
    • one year ago
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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.

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

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

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

  10. mathcalculus
    • one year ago
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    ok show it.

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

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

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

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

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

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

  17. calmat01
    • one year ago
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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.

  18. calmat01
    • one year ago
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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.

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

  20. calmat01
    • one year ago
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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?

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

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

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

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

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

  26. calmat01
    • one year ago
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    There you go.

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

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

  29. calmat01
    • one year ago
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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.

  30. mathcalculus
    • one year ago
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    O_o

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

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

  33. calmat01
    • one year ago
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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.

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

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

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

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

  38. mathcalculus
    • one year ago
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    @calmat01

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

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

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

  42. calmat01
    • one year ago
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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.

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

  44. mathcalculus
    • one year ago
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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

  45. dan815
    • one year ago
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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)

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

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

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

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

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

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

  52. dan815
    • one year ago
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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|

  53. dan815
    • one year ago
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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

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

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

  56. dan815
    • one year ago
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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|

  57. dan815
    • one year ago
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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

  58. dan815
    • one year ago
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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

  59. dan815
    • one year ago
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    make sense now?

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

  61. dan815
    • one year ago
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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

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

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

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

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