anonymous
  • anonymous
At what value of x does the local max of f(x) occur?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[f(x)=\int\limits_{0}^{x}\frac{ t^2-1 }{ 1+\cos^2(t) }dt\]
anonymous
  • anonymous
So they gave you the derivative... When it is 0?
anonymous
  • anonymous
Also, when is it undefined?

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anonymous
  • anonymous
It seems it is always defined.
anonymous
  • anonymous
so it's zero when x=0, right?
zepdrix
  • zepdrix
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,\[\large f'(x)=\frac{x^2-1}{1+\cos^2(x)}\]
anonymous
  • anonymous
Ok yeah I'm following that now
zepdrix
  • zepdrix
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.
anonymous
  • anonymous
so the critical points are -1 and 1
zepdrix
  • zepdrix
cool :) Remember how to determine if each point is a max or min?
anonymous
  • anonymous
using the 2nd derivative test so the max is at -1
zepdrix
  • zepdrix
Yah that's what I'm coming up with also. Cool!
anonymous
  • anonymous
THANKS!
zepdrix
  • zepdrix
heh np c:

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