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mathcalculus

  • 2 years ago

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 ??

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  1. mattt9
    • 2 years ago
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    is the x^2+7 all under the -4x

  2. mathcalculus
    • 2 years ago
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    yup

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

  4. mathcalculus
    • 2 years ago
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    @mattt9

  5. mattt9
    • 2 years ago
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    and what was your derivative again?

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

  7. mattt9
    • 2 years ago
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    are you sure that is correct wouldn't the derivative of the top be -4 not 1

  8. mathcalculus
    • 2 years ago
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    yeah explain that please

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

  10. mattt9
    • 2 years ago
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    |dw:1363315804166:dw|

  11. mattt9
    • 2 years ago
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    you still use the constant if it's attached to the x like that.

  12. mattt9
    • 2 years ago
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    (d/dx) of 2x is 2 right?

  13. mathcalculus
    • 2 years ago
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    oh so when you add you don't use the constant right?

  14. mattt9
    • 2 years ago
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    as a random example

  15. mattt9
    • 2 years ago
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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.

  16. mathcalculus
    • 2 years ago
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    oh so you multiplied them which is -4x... then just fnd the derivative right>

  17. mathcalculus
    • 2 years ago
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    for example if it is -4+x then the derivative is 1.

  18. mathcalculus
    • 2 years ago
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    but if (-4)*x you keep them as -4 and find the derivative which is -4

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

  20. mathcalculus
    • 2 years ago
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    awesome so let me do that.

  21. mattt9
    • 2 years ago
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    okay

  22. mathcalculus
    • 2 years ago
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    how did you get 4x^2-28?

  23. mathcalculus
    • 2 years ago
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    what happened with the -6x?

  24. mattt9
    • 2 years ago
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    what -6x

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

  26. mattt9
    • 2 years ago
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    |dw:1363317471013:dw|

  27. mattt9
    • 2 years ago
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    that's the numerator

  28. mathcalculus
    • 2 years ago
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    yeah but you wrote.. (4x2−28)/(x2+7)(x2+7)

  29. mathcalculus
    • 2 years ago
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    ^**

  30. mattt9
    • 2 years ago
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    yeah i know. but you had problems with the numerator. so i just showed you how i got the numerator

  31. mattt9
    • 2 years ago
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    that is the full derivative.

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

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

  34. mattt9
    • 2 years ago
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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)

  35. mattt9
    • 2 years ago
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    at this point you will see that x = +-root(7) = +- 2.646

  36. mattt9
    • 2 years ago
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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.

  37. mathcalculus
    • 2 years ago
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    matt you make sense. are you there @mattt9

  38. mathcalculus
    • 2 years ago
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    so we don't test the critical points correct?

  39. mattt9
    • 2 years ago
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    those are the critical points.

  40. mathcalculus
    • 2 years ago
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    a) because the intervals are [1,4] and b) intervals are from [1,5]

  41. mattt9
    • 2 years ago
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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

  42. mathcalculus
    • 2 years ago
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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...

  43. mathcalculus
    • 2 years ago
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    @mattt9 i would appreciate the help very much.

  44. mattt9
    • 2 years ago
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    ummmmm what do you get when you plug in 2.626 into the fn.?

  45. mathcalculus
    • 2 years ago
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    2.626 into the fn.?

  46. mathcalculus
    • 2 years ago
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    the radical 7 or negative radical 7 is 2.64575

  47. mattt9
    • 2 years ago
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    (-4)(2.626)/(2.626^2+7) = -0.7559

  48. mattt9
    • 2 years ago
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    that should be your minimum for both.

  49. mathcalculus
    • 2 years ago
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    youre right, i didn't see the negative by mistake.

  50. mathcalculus
    • 2 years ago
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    thank you!!

  51. mattt9
    • 2 years ago
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    you areee welcome !

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