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heradog

  • 3 years ago

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

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

  2. wio
    • 3 years ago
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    So they gave you the derivative... When it is 0?

  3. wio
    • 3 years ago
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    Also, when is it undefined?

  4. wio
    • 3 years ago
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    It seems it is always defined.

  5. heradog
    • 3 years ago
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    so it's zero when x=0, right?

  6. zepdrix
    • 3 years ago
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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,\[\large f'(x)=\frac{x^2-1}{1+\cos^2(x)}\]

  7. heradog
    • 3 years ago
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    Ok yeah I'm following that now

  8. zepdrix
    • 3 years ago
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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.

  9. heradog
    • 3 years ago
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    so the critical points are -1 and 1

  10. zepdrix
    • 3 years ago
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    cool :) Remember how to determine if each point is a max or min?

  11. heradog
    • 3 years ago
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    using the 2nd derivative test so the max is at -1

  12. zepdrix
    • 3 years ago
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    Yah that's what I'm coming up with also. Cool!

  13. heradog
    • 3 years ago
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    THANKS!

  14. zepdrix
    • 3 years ago
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    heh np c:

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