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

wio
 one year ago
Best ResponseYou've already chosen the best response.0So they gave you the derivative... When it is 0?

wio
 one year ago
Best ResponseYou've already chosen the best response.0Also, when is it undefined?

wio
 one year ago
Best ResponseYou've already chosen the best response.0It seems it is always defined.

heradog
 one year ago
Best ResponseYou've already chosen the best response.0so it's zero when x=0, right?

zepdrix
 one year 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,\[\large f'(x)=\frac{x^21}{1+\cos^2(x)}\]

heradog
 one year ago
Best ResponseYou've already chosen the best response.0Ok yeah I'm following that now

zepdrix
 one year 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.

heradog
 one year ago
Best ResponseYou've already chosen the best response.0so the critical points are 1 and 1

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

heradog
 one year ago
Best ResponseYou've already chosen the best response.0using the 2nd derivative test so the max is at 1

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