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anonymous
 one year ago
Find the interval on which the curve of "picture below" is concave up
anonymous
 one year ago
Find the interval on which the curve of "picture below" is concave up

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Ashleyisakitty @ganeshie8 @Hero @jim_thompson5910

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know i have to find the second derivative im just not sure how

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=\int\limits_{0}^{x}\frac{ dt }{ 1+t+t^2 }=\int\limits_{0}^{x}\frac{ dt }{ t^2+t+\frac{ 1 }{ 4 }\frac{ 1 }{ 4 }+1 }\] \[=\int\limits_{0}^{x}\frac{ dt }{ \left( t+\frac{ 1 }{ 2 } \right)^2+\frac{ 3 }{ 4 } }=\int\limits_{0}^{x}\frac{ dt }{ \left( t+\frac{ 1 }{ 2 } \right)^2+\left( \frac{ \sqrt{3} }{ 2 } \right)^2 }\] \[=\frac{ 1 }{ \frac{ \sqrt{3} }{ 2 } }\tan^{1} \frac{ t+\frac{ 1 }{ 2 } }{ \frac{ \sqrt{3} }{ 2 } }0\rightarrow x\] \[=\frac{ 2 }{ \sqrt{3} }\tan^{1} \frac{ 2t+1 }{ \sqrt{3} }0\rightarrow\ x\] \[=\frac{ 2 }{ \sqrt{3} }\left\{ \tan^{1} \left( \frac{ 2x+1 }{ \sqrt{3} }\right)\tan^{1}\frac{ 1 }{ \sqrt{3} } \right\}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow okay give me a second to read that...

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3@surjithayer there's no need to integrate you just need to find \(\Large \frac{d^2y}{dx^2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh no first derivative is the integrand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh what @jim_thompson5910 said

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3the first derivative is not too bad because you can use the fundamental theorem of calculus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea thats what im having a problem with , do i just plug in x for every where t is at and then integrate that or do i just plug in x and then that's my first derivative?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first derivative replace each \(t\) in the integrand by an \(x\) that is all then take the derivative of the result to get the second derivative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0interval is 0 to x or we can change it to x then diff.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean derive not integrate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah plug in x for t

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then DIFFERENTIATE that , not integrate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay okay one second let me do that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i also said differentiate.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3Fundamental Theorem of Calculus (one of the 2 parts) If \[\Large g(x) = \int_{a}^{x}f(t)dt\] then \[\Large g \ '(x) = f(x)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3surjithayer you weren't incorrect. It was just a step in the opposite direction adding unnecessary steps. Then again, doing it that way helps see how the FTC works

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, i have not used that theorem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you for reminding me.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay im getting \[\frac{(2x+1) }{ (x^2+x+1)^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i would have written the numerator as \[2x1\] so that you can solve \[2x1>0\] rather easily

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay out of curiosity whats the second part of the theorem ... it sounds very "fundamental "

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is that \(F(b)F(a)\) business

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3Other part of FTC \[\Large \int_{a}^{b}f \ '(x) dx = f(b)  f(a)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0arh damn i don't know who to give the medal too you were all so helpful

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3Most books make F ' = f, which is confusing to me because there are 2 different f's to keep track of

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats weird every time i see F it means \[intf(x)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3F ' = f is equivalent to saying "F is the indefinite integral of f"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0notation like to be the death of us all nothing worse than having two identical notations for different things

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ill keep that in mind :D thanks again every one !!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3the reason I set up FTC 2 that way was because most calc courses start with derivatives and then teach integrals. So why not set up the integral in terms of a derivative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay real quick so i should graph that and where its positive is the integral that the original graph is concave up right?!?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is concave up where \[2x1>0\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3Exactly. The denominator is always positive, so you can ignore that portion and focus on where the numerator is positive.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it would be (infinity , .5) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay ill go ahead and take that as a silent nod

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3yes when x < 1/2
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