Find the interval on which the curve of "picture below" is concave up

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Find the interval on which the curve of "picture below" is concave up

Mathematics
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i know i have to find the second derivative im just not sure how

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\[y=\int\limits_{0}^{x}\frac{ dt }{ 1+t+t^2 }=\int\limits_{0}^{x}\frac{ dt }{ t^2+t+\frac{ 1 }{ 4 }-\frac{ 1 }{ 4 }+1 }\] \[=\int\limits_{0}^{x}\frac{ dt }{ \left( t+\frac{ 1 }{ 2 } \right)^2+\frac{ 3 }{ 4 } }=\int\limits_{0}^{x}\frac{ dt }{ \left( t+\frac{ 1 }{ 2 } \right)^2+\left( \frac{ \sqrt{3} }{ 2 } \right)^2 }\] \[=\frac{ 1 }{ \frac{ \sqrt{3} }{ 2 } }\tan^{-1} \frac{ t+\frac{ 1 }{ 2 } }{ \frac{ \sqrt{3} }{ 2 } }|0\rightarrow x\] \[=\frac{ 2 }{ \sqrt{3} }\tan^{-1} \frac{ 2t+1 }{ \sqrt{3} }|0\rightarrow\ x\] \[=\frac{ 2 }{ \sqrt{3} }\left\{ \tan^{-1} \left( \frac{ 2x+1 }{ \sqrt{3} }\right)-\tan^{-1}\frac{ 1 }{ \sqrt{3} } \right\}\]
wow okay give me a second to read that...
@surjithayer there's no need to integrate you just need to find \(\Large \frac{d^2y}{dx^2}\)
oh no first derivative is the integrand
oh what @jim_thompson5910 said
the first derivative is not too bad because you can use the fundamental theorem of calculus
yea 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?
first 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
interval is 0 to x or we can change it to x then diff.
i mean derive not integrate
yeah plug in x for t
then DIFFERENTIATE that , not integrate
okay okay one second let me do that
i also said differentiate.
Fundamental Theorem of Calculus (one of the 2 parts) If \[\Large g(x) = \int_{a}^{x}f(t)dt\] then \[\Large g \ '(x) = f(x)\]
surjithayer 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
sorry, i have not used that theorem.
thank you for reminding me.
okay im getting \[\frac{-(2x+1) }{ (x^2+x+1)^2}\]
looks good to me
agreed
i would have written the numerator as \[-2x-1\] so that you can solve \[-2x-1>0\] rather easily
okay out of curiosity whats the second part of the theorem ... it sounds very "fundamental "
that is that \(F(b)-F(a)\) business
Other part of FTC \[\Large \int_{a}^{b}f \ '(x) dx = f(b) - f(a)\]
arh damn i don't know who to give the medal too you were all so helpful
Most books make F ' = f, which is confusing to me because there are 2 different f's to keep track of
thats weird every time i see F it means \[intf(x)\]
F ' = f is equivalent to saying "F is the indefinite integral of f"
notation like to be the death of us all nothing worse than having two identical notations for different things
ill keep that in mind :D thanks again every one !!
the 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
no problem
okay real quick so i should graph that and where its positive is the integral that the original graph is concave up right?!?
it is concave up where \[-2x-1>0\]
Exactly. The denominator is always positive, so you can ignore that portion and focus on where the numerator is positive.
so it would be (-infinity , -.5) right?
okay ill go ahead and take that as a silent nod
yes when x < -1/2

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