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finaynay

  • 2 years ago

Find the point(s) guaranteed by the mean value theorem for f(x)=|x^2-9| over the interval [0,2]

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  1. math>philosophy
    • 2 years ago
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    x^2 - 0?

  2. finaynay
    • 2 years ago
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    i meant 9 lol

  3. math>philosophy
    • 2 years ago
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    lol

  4. math>philosophy
    • 2 years ago
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    So what are the conditions for the MVT

  5. finaynay
    • 2 years ago
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    well it is already guaranteed that the mean value theorem applies so i did f'(x)= f(2)-f(0)/2 f'(x)=-2 then what do i find the derivative of f(x)=|x^2-9| because i dont know how to find the derivative of a absolute value

  6. math>philosophy
    • 2 years ago
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    \[\sqrt{x^2} = |x|\]

  7. finaynay
    • 2 years ago
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    how would i do that for |x^2-9| and did i do the beginning of the problem correctly

  8. Callisto
    • 2 years ago
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    y =|x^2-9| Let u =x^2-9 => \(y=\sqrt{ u^2}\) \[\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}\] Does that help?

  9. finaynay
    • 2 years ago
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    i still don't understand

  10. Callisto
    • 2 years ago
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    Which part?

  11. finaynay
    • 2 years ago
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    how i'm supposed to take the derivative of \[|x ^{2}-9|\]

  12. Callisto
    • 2 years ago
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    Do you know that \(\sqrt{a^2} = |a| \)?

  13. finaynay
    • 2 years ago
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    yes but how do i apply that \[\sqrt{(x-9)^{2}}\] isn't the derivative

  14. Callisto
    • 2 years ago
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    It's the question :P To find the derivative of |x^2 -9|, you can find the derivative of \(\sqrt{(x^2-9)^2}\), which is the same as |x^2 -9|. Btw, it should be \(\sqrt{(x^2-9)^2}\), not \(\sqrt{(x-9)^2}\)

  15. finaynay
    • 2 years ago
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    oh okkkkkkk but is that even what i'm supposed to be doing in this problem

  16. Callisto
    • 2 years ago
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    You need to find f'(x), so, I think so :|

  17. finaynay
    • 2 years ago
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    for realsssss noooooooo

  18. Callisto
    • 2 years ago
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    Hmmm.. MVT: Suppose y=f(x) is continuous on a closed interval [a,b] and differentiable on the interior (a,b). Then there is at least one point c at which \(\frac{f(b)-f(a)}{b-a} = f'(c)\) In this question, the conditions are satisfied, that is f(x) = |x^2-9| is continuous on the closed interval [0, 2] and is differentiable on (0, 2) So, just apply it: \(\frac{f(2)-f(0)}{2} = f'(c)\) And you have to find f'(x), isn't it?

  19. finaynay
    • 2 years ago
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    yeah

  20. Callisto
    • 2 years ago
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    So... just differentiate that guy :|

  21. finaynay
    • 2 years ago
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    so the derivative is 2x

  22. finaynay
    • 2 years ago
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    x=-1?

  23. satellite73
    • 2 years ago
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    not quite

  24. satellite73
    • 2 years ago
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    on the interval \([0,2]\) you have \(|x^2-9|=9-x^2\)

  25. finaynay
    • 2 years ago
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    how is it 9−x^2

  26. satellite73
    • 2 years ago
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    since this is a quadratic, the "point guaranteed to exist by the mean values theorem will be in the midpoint of the interval, namely at \(x=1\)

  27. satellite73
    • 2 years ago
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    if \(-3<x<3\) we have \(x^2-9<0\) and so \(|x^2-9|=-(x^2-9)=9-x^2\)

  28. satellite73
    • 2 years ago
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    in english \(x^2-9\) is negative on the interval \([0,2]\) and so the absolute value is minus the whole thing , namely \(9-x^2\)

  29. finaynay
    • 2 years ago
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    okay so the derivative is -2x right and then you set that equal to the mean and you solve for x and you get x=1?

  30. satellite73
    • 2 years ago
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    lets try it

  31. satellite73
    • 2 years ago
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    \(f(0)=9\) and \(f(2)=5\) and so \(\frac{f(2)-f(0)}{2-0}=\frac{5-9}{2}=-4\)

  32. finaynay
    • 2 years ago
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    isn't it -2

  33. satellite73
    • 2 years ago
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    ooops bad arithmetic

  34. finaynay
    • 2 years ago
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    oh okay lol

  35. math>philosophy
    • 2 years ago
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    <3

  36. satellite73
    • 2 years ago
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    yes it is \(-2\) now we set \(-2x=-2\) and guess what? \(x=1\)

  37. finaynay
    • 2 years ago
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    -2=f'(x)

  38. finaynay
    • 2 years ago
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    yessss! alright i get it now

  39. satellite73
    • 2 years ago
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    you can convince yourself, if you have the time in inclination, that the "point guaranteed to exist by the mean value theorem" is smack in the middle of the interval if your function is a quadratic

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