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heradog Group Title

At what value of x does the local max of f(x) occur?

  • one year ago
  • one year ago

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  1. heradog Group Title
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    \[f(x)=\int\limits_{0}^{x}\frac{ t^2-1 }{ 1+\cos^2(t) }dt\]

    • one year ago
  2. wio Group Title
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    So they gave you the derivative... When it is 0?

    • one year ago
  3. wio Group Title
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    Also, when is it undefined?

    • one year ago
  4. wio Group Title
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    It seems it is always defined.

    • one year ago
  5. heradog Group Title
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    so it's zero when x=0, right?

    • one year ago
  6. zepdrix Group Title
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    Ummmmmm I think what we want to do is use the `Fundamental Theorem of Calculus, Part 1` for this problem. Which states,\[\large \frac{d}{dx} \int\limits_0^{x}f(t)dt \qquad = \qquad f(x)\] For this problem, we're given \(\large f(x)\). We can find a maximum by first finding critical points the function. Remember how to do that? Critical points occur when \(\large f'(x)=0\). \[\large f(x)=\int\limits\limits_{0}^{x}\frac{ t^2-1 }{ 1+\cos^2(t) }dt\] Applying the FTC, Part 1 gives us,\[\large f'(x)=\frac{x^2-1}{1+\cos^2(x)}\]

    • one year ago
  7. heradog Group Title
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    Ok yeah I'm following that now

    • one year ago
  8. zepdrix Group Title
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    Setting it equal to zero, \[\large 0=\frac{x^2-1}{1+\cos^2(x)}\] We should be able to find some critical points, and then by some method we can determine which point(s) are max/min or whatever.

    • one year ago
  9. heradog Group Title
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    so the critical points are -1 and 1

    • one year ago
  10. zepdrix Group Title
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    cool :) Remember how to determine if each point is a max or min?

    • one year ago
  11. heradog Group Title
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    using the 2nd derivative test so the max is at -1

    • one year ago
  12. zepdrix Group Title
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    Yah that's what I'm coming up with also. Cool!

    • one year ago
  13. heradog Group Title
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    THANKS!

    • one year ago
  14. zepdrix Group Title
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    heh np c:

    • one year ago
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