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

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

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

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

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

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

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

zepdrixBest ResponseYou've already chosen the best response.1
Setting 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.
 one year ago

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

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

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

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