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anonymous
 3 years ago
At what value of x does the local max of f(x) occur?
anonymous
 3 years ago
At what value of x does the local max of f(x) occur?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=\int\limits_{0}^{x}\frac{ t^21 }{ 1+\cos^2(t) }dt\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So they gave you the derivative... When it is 0?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Also, when is it undefined?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It seems it is always defined.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it's zero when x=0, right?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Ummmmmm 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 \(\large f'(x)=0\). \[\large f(x)=\int\limits\limits_{0}^{x}\frac{ t^21 }{ 1+\cos^2(t) }dt\] Applying the FTC, Part 1 gives us,\[\large f'(x)=\frac{x^21}{1+\cos^2(x)}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok yeah I'm following that now

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Setting it equal to zero, \[\large 0=\frac{x^21}{1+\cos^2(x)}\] 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
 3 years ago
Best ResponseYou've already chosen the best response.0so the critical points are 1 and 1

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1cool :) Remember how to determine if each point is a max or min?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0using the 2nd derivative test so the max is at 1

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Yah that's what I'm coming up with also. Cool!
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