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Dido525
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Find the intervals on which the function f(x) is increasing.
 one year ago
 one year ago
Dido525 Group Title
Find the intervals on which the function f(x) is increasing.
 one year ago
 one year ago

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Dido525 Group TitleBest ResponseYou've already chosen the best response.0
dw:1353112573724:dw
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix . I believe I would find the derivative using FTC and then antiderivative that to find f(x) right?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
I think you're half right, I think you want to do the first half of what you said.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large f(x)=\int\limits_{0}^{x}(1+t^2)e^{t^2}dt\] \[\huge f'(x)=(1+x^2)e^{x^2}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Then just find critical points to find your intervals... i think :O
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Yeah, but wouldn't I also need to know what f(x) is?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Actually... I wouldn't right? All I need is the derivative.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Yah I don't think they want us to SOLVE for f(x). They've given us f(x), it just looks a little funny. We simply want intervals of increasing/decreasing. So we only need to deal with f'(x).
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Thanks :) .
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Guess it's my OCD telling me to find f(x) .
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Hmm i could be wrong, but I don't think you can actually solve that integral using elementary methods :D heh
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
because of the e^t^2 term :o
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Yeah, guess so.
 one year ago
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