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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]

  • one year ago
  • one year ago

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

    • one year ago
  2. finaynay Group Title
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    i meant 9 lol

    • one year ago
  3. math>philosophy Group Title
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    lol

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

    • one year 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

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

    • one year 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

    • one year 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?

    • one year ago
  9. finaynay Group Title
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    i still don't understand

    • one year ago
  10. Callisto Group Title
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    Which part?

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

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

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

    • one year 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 \(\sqrt{(x-9)^2}\)

    • one year 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

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

    • one year ago
  17. finaynay Group Title
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    for realsssss noooooooo

    • one year 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?

    • one year ago
  19. finaynay Group Title
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    yeah

    • one year ago
  20. Callisto Group Title
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    So... just differentiate that guy :|

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

    • one year ago
  22. finaynay Group Title
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    x=-1?

    • one year ago
  23. satellite73 Group Title
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    not quite

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

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

    • one year 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 \(x=1\)

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

    • one year 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 \(9-x^2\)

    • one year 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?

    • one year ago
  30. satellite73 Group Title
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    lets try it

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

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

    • one year ago
  33. satellite73 Group Title
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    ooops bad arithmetic

    • one year ago
  34. finaynay Group Title
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    oh okay lol

    • one year ago
  35. math>philosophy Group Title
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    <3

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

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

    • one year ago
  38. finaynay Group Title
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    yessss! alright i get it now

    • one year 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

    • one year ago
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