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mathcalculus Group Title

help help help!!!! Determine the extrema of f(x)= (-4)* x/ x^2+7 below on the given interval (a) on [1,4] The minimum is ?? and the maximum is ?? (b) on [1,5] The minimum is ?? and the maximum is ??

  • one year ago
  • one year ago

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  1. mattt9 Group Title
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    is the x^2+7 all under the -4x

    • one year ago
  2. mathcalculus Group Title
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    yup

    • one year ago
  3. mathcalculus Group Title
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    yup\[ \frac{ (-4)*x}{ x^{2}+7}\]

    • one year ago
  4. mathcalculus Group Title
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    @mattt9

    • one year ago
  5. mattt9 Group Title
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    and what was your derivative again?

    • one year ago
  6. mathcalculus Group Title
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    \[\frac{ 1*(x ^{2}+ 7)-2x(-4)*x}{ (x^{2}+7) ^2 }\]

    • one year ago
  7. mattt9 Group Title
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    are you sure that is correct wouldn't the derivative of the top be -4 not 1

    • one year ago
  8. mathcalculus Group Title
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    yeah explain that please

    • one year ago
  9. mathcalculus Group Title
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    maybe I did get it wrong. but i assumed since -4 is a constant i didn't use it...

    • one year ago
  10. mattt9 Group Title
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    |dw:1363315804166:dw|

    • one year ago
  11. mattt9 Group Title
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    you still use the constant if it's attached to the x like that.

    • one year ago
  12. mattt9 Group Title
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    (d/dx) of 2x is 2 right?

    • one year ago
  13. mathcalculus Group Title
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    oh so when you add you don't use the constant right?

    • one year ago
  14. mattt9 Group Title
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    as a random example

    • one year ago
  15. mattt9 Group Title
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    you wouldn't use the 7 when determining the derivative of the bottom since it is not attached to an x value.

    • one year ago
  16. mathcalculus Group Title
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    oh so you multiplied them which is -4x... then just fnd the derivative right>

    • one year ago
  17. mathcalculus Group Title
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    for example if it is -4+x then the derivative is 1.

    • one year ago
  18. mathcalculus Group Title
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    but if (-4)*x you keep them as -4 and find the derivative which is -4

    • one year ago
  19. mattt9 Group Title
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    right. then your derivative simplifies to : \[\frac{ (4x^2-28) }{ (x^2+7)(x^2+7) }\]

    • one year ago
  20. mathcalculus Group Title
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    awesome so let me do that.

    • one year ago
  21. mattt9 Group Title
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    okay

    • one year ago
  22. mathcalculus Group Title
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    how did you get 4x^2-28?

    • one year ago
  23. mathcalculus Group Title
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    what happened with the -6x?

    • one year ago
  24. mattt9 Group Title
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    what -6x

    • one year ago
  25. mathcalculus Group Title
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    i thought the derivative would be -4x^2-28-6x on the numerator

    • one year ago
  26. mattt9 Group Title
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    |dw:1363317471013:dw|

    • one year ago
  27. mattt9 Group Title
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    that's the numerator

    • one year ago
  28. mathcalculus Group Title
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    yeah but you wrote.. (4x2−28)/(x2+7)(x2+7)

    • one year ago
  29. mathcalculus Group Title
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    ^**

    • one year ago
  30. mattt9 Group Title
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    yeah i know. but you had problems with the numerator. so i just showed you how i got the numerator

    • one year ago
  31. mattt9 Group Title
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    that is the full derivative.

    • one year ago
  32. mattt9 Group Title
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    anyways. you end up equation that to zero. \[\frac{ (4x^2-28) }{ (x^2+7)(x^2+7) }=0\]

    • one year ago
  33. mattt9 Group Title
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    the (x^2+7) cancels out and you are left with 4x^2-28 = 0

    • one year ago
  34. mattt9 Group Title
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    then solve for x to obtain a value for which the slope of the function is zero (a possible max or min)

    • one year ago
  35. mattt9 Group Title
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    at this point you will see that x = +-root(7) = +- 2.646

    • one year ago
  36. mattt9 Group Title
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    then you see that x = -2.626 is not in your given interval so you would not test this value. in the first interval of [1,4] you would test the x-values of x = 1 x = 4 and your found x value of x = 2.626 out of these three numbers one will be highest and one will be lowest, those should be your max and min for the function respectively. Let me know if any of that is incorrect.

    • one year ago
  37. mathcalculus Group Title
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    matt you make sense. are you there @mattt9

    • one year ago
  38. mathcalculus Group Title
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    so we don't test the critical points correct?

    • one year ago
  39. mattt9 Group Title
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    those are the critical points.

    • one year ago
  40. mathcalculus Group Title
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    a) because the intervals are [1,4] and b) intervals are from [1,5]

    • one year ago
  41. mattt9 Group Title
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    or at least 2.626 is for the function. 1 and 4 are for the interval. or maybe that terminology isn't used i can't remember but you would test all three of those points

    • one year ago
  42. mathcalculus Group Title
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    i did test them all out. i wrote (a) on [1,4] The minimum is -.695652 and the maximum is -1/2 (b) on [1,5] The minimum is -.625 and the maximum is -1/2 but when i click submit it says one of them is wrong...

    • one year ago
  43. mathcalculus Group Title
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    @mattt9 i would appreciate the help very much.

    • one year ago
  44. mattt9 Group Title
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    ummmmm what do you get when you plug in 2.626 into the fn.?

    • one year ago
  45. mathcalculus Group Title
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    2.626 into the fn.?

    • one year ago
  46. mathcalculus Group Title
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    the radical 7 or negative radical 7 is 2.64575

    • one year ago
  47. mattt9 Group Title
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    (-4)(2.626)/(2.626^2+7) = -0.7559

    • one year ago
  48. mattt9 Group Title
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    that should be your minimum for both.

    • one year ago
  49. mathcalculus Group Title
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    youre right, i didn't see the negative by mistake.

    • one year ago
  50. mathcalculus Group Title
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    thank you!!

    • one year ago
  51. mattt9 Group Title
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    you areee welcome !

    • one year ago
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