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zeesbrat3

  • one year ago

If f(x) = |(x2 − 4)(x2 + 2)|, how many numbers in the interval [0, 1] satisfy the conclusion of the Mean Value Theorem?

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  1. zeesbrat3
    • one year ago
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    @Astrophysics @ganeshie8 @Michele_Laino

  2. zeesbrat3
    • one year ago
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    help..

  3. Michele_Laino
    • one year ago
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    I'm pondering...

  4. zeesbrat3
    • one year ago
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    Oh, sorry

  5. Michele_Laino
    • one year ago
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    we have to note that we have the subsequent inequality: \[\Large {x^2} - 4 < 0,\quad {\text{if }} - 2 < x < 2\]

  6. Michele_Laino
    • one year ago
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    am I right?

  7. zeesbrat3
    • one year ago
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    honestly, im confused

  8. Michele_Laino
    • one year ago
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    why?

  9. Michele_Laino
    • one year ago
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    it is a simple inequality

  10. zeesbrat3
    • one year ago
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    no i understand the inequality, im confused by how you got it

  11. Michele_Laino
    • one year ago
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    ok! I wrote that inequality, because inside the interval [0,1] your function can be rewritten as follows: \[\Large f\left( x \right) = - \left( {{x^2} - 4} \right)\left( {{x^2} + 2} \right) = \left( {4 - {x^2}} \right)\left( {{x^2} + 2} \right)\]

  12. zeesbrat3
    • one year ago
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    otherwise any other would be out of bounds

  13. Michele_Laino
    • one year ago
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    no, it is a step in order to get rid the absolute value symbol

  14. zeesbrat3
    • one year ago
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    oh, okay.

  15. Michele_Laino
    • one year ago
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    now, we can apply the theorem af mean value

  16. Michele_Laino
    • one year ago
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    so, we have to compute this: \[\Large \frac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}} = \frac{{3 \times 3 - 4 \times 2}}{1} = ...\]

  17. zeesbrat3
    • one year ago
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    \[f'(c) = \frac{ f(b) - f(a) }{ b - a }\]

  18. zeesbrat3
    • one year ago
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    1

  19. Michele_Laino
    • one year ago
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    are you sure?

  20. Michele_Laino
    • one year ago
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    \[\large \frac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}} = \frac{{3 \times 3 - 4 \times 2}}{1} = \frac{{6 - 8}}{1} = ...?\]

  21. zeesbrat3
    • one year ago
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    isnt 3x3 9?

  22. Michele_Laino
    • one year ago
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    sorry I have made a big error, you are right

  23. zeesbrat3
    • one year ago
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    it's okay :)

  24. Michele_Laino
    • one year ago
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    sorry again!!

  25. zeesbrat3
    • one year ago
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    no apologies necessary! thank you for your help

  26. Michele_Laino
    • one year ago
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    next we have to compute the first derivative at a generic point x, namely we have to compute the first derivative of this function: \[\Large f\left( x \right) = \left( {4 - {x^2}} \right)\left( {{x^2} + 2} \right)\]

  27. Michele_Laino
    • one year ago
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    what do you get?

  28. zeesbrat3
    • one year ago
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    \[-4x(x^2-1)\]

  29. Michele_Laino
    • one year ago
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    correct!

  30. Michele_Laino
    • one year ago
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    then we have to solve this equation: \[ \Large - 4x\left( {{x^2} - 1} \right) = 1\]

  31. Michele_Laino
    • one year ago
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    it is not a simple equation

  32. zeesbrat3
    • one year ago
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    \[x^3 = -\frac{ 1 }{ 2 }\] ?

  33. Michele_Laino
    • one year ago
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    how did you do to write that expression?

  34. zeesbrat3
    • one year ago
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    \[-4x(x^2-1) = 1\]\[\frac{ 1 }{ -4x } = x^2\] \[\frac{ 1 }{ -4x } +1 = x^2\] \[1+1 = -4x^3\] \[- \frac{ 1 }{ 2 } = x^3\]

  35. Michele_Laino
    • one year ago
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    I don't understand this step: \[\frac{1}{{ - 4x}} = {x^2}\]

  36. Michele_Laino
    • one year ago
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    anyway it is not a solution of our equation

  37. zeesbrat3
    • one year ago
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    Sorry, i forgot the -1

  38. zeesbrat3
    • one year ago
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    How would you get the proper solution?

  39. Michele_Laino
    • one year ago
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    I questioned wolfram alpha, here are the solutions:

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  40. zeesbrat3
    • one year ago
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    So, 3 solutions in the interval

  41. Michele_Laino
    • one year ago
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    we have two solutions, inside the interval [0,1]

  42. zeesbrat3
    • one year ago
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    Whoops, my bad! I looked outside the interval

  43. Michele_Laino
    • one year ago
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  44. Michele_Laino
    • one year ago
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    do you see them?

  45. zeesbrat3
    • one year ago
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    Yes. Since one of them is in the negatives, it is not in the interval and does not matter to answer the question

  46. Michele_Laino
    • one year ago
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    yes! that's right!

  47. zeesbrat3
    • one year ago
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    Thank you!

  48. Michele_Laino
    • one year ago
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    :)

  49. zeesbrat3
    • one year ago
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    I tried doing the same thing on this problem, but I keep getting no solutions: A particle moves along the x-axis with position function s(t) = xe^x. How many times in the interval [−5, 5] is the velocity equal to 0?

  50. Loser66
    • one year ago
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    velocity = (s(t))' take derivative and let it =0 to solve for x .

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