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finaynay Group Title

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

  • 2 years ago
  • 2 years ago

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  1. math>philosophy Group Title
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    x^2 - 0?

    • 2 years ago
  2. finaynay Group Title
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    i meant 9 lol

    • 2 years ago
  3. math>philosophy Group Title
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    lol

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

    • 2 years ago
  5. finaynay Group Title
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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

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

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

    • 2 years ago
  8. Callisto Group Title
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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?

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

    • 2 years ago
  10. Callisto Group Title
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    Which part?

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

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

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

    • 2 years ago
  14. Callisto Group Title
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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 \(\sqrt{(x-9)^2}\)

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

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

    • 2 years ago
  17. finaynay Group Title
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    for realsssss noooooooo

    • 2 years ago
  18. Callisto Group Title
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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?

    • 2 years ago
  19. finaynay Group Title
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    yeah

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

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

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

    • 2 years ago
  23. satellite73 Group Title
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    not quite

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

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

    • 2 years ago
  26. satellite73 Group Title
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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 \(x=1\)

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

    • 2 years ago
  28. satellite73 Group Title
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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 \(9-x^2\)

    • 2 years ago
  29. finaynay Group Title
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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?

    • 2 years ago
  30. satellite73 Group Title
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    lets try it

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

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

    • 2 years ago
  33. satellite73 Group Title
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    ooops bad arithmetic

    • 2 years ago
  34. finaynay Group Title
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    oh okay lol

    • 2 years ago
  35. math>philosophy Group Title
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    <3

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

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

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

    • 2 years ago
  39. satellite73 Group Title
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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

    • 2 years ago
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