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

- anonymous

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

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- anonymous

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- anonymous

@Ashleyisakitty @ganeshie8 @Hero @jim_thompson5910

- anonymous

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

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## More answers

- anonymous

\[y=\int\limits_{0}^{x}\frac{ dt }{ 1+t+t^2 }=\int\limits_{0}^{x}\frac{ dt }{ t^2+t+\frac{ 1 }{ 4 }-\frac{ 1 }{ 4 }+1 }\]
\[=\int\limits_{0}^{x}\frac{ dt }{ \left( t+\frac{ 1 }{ 2 } \right)^2+\frac{ 3 }{ 4 } }=\int\limits_{0}^{x}\frac{ dt }{ \left( t+\frac{ 1 }{ 2 } \right)^2+\left( \frac{ \sqrt{3} }{ 2 } \right)^2 }\]
\[=\frac{ 1 }{ \frac{ \sqrt{3} }{ 2 } }\tan^{-1} \frac{ t+\frac{ 1 }{ 2 } }{ \frac{ \sqrt{3} }{ 2 } }|0\rightarrow x\]
\[=\frac{ 2 }{ \sqrt{3} }\tan^{-1} \frac{ 2t+1 }{ \sqrt{3} }|0\rightarrow\ x\]
\[=\frac{ 2 }{ \sqrt{3} }\left\{ \tan^{-1} \left( \frac{ 2x+1 }{ \sqrt{3} }\right)-\tan^{-1}\frac{ 1 }{ \sqrt{3} } \right\}\]

- anonymous

wow okay give me a second to read that...

- jim_thompson5910

@surjithayer there's no need to integrate
you just need to find \(\Large \frac{d^2y}{dx^2}\)

- anonymous

oh no
first derivative is the integrand

- anonymous

oh what @jim_thompson5910 said

- jim_thompson5910

the first derivative is not too bad because you can use the fundamental theorem of calculus

- anonymous

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?

- anonymous

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

- anonymous

interval is 0 to x or we can change it to x then diff.

- anonymous

i mean derive not integrate

- anonymous

yeah plug in x for t

- anonymous

then DIFFERENTIATE that , not integrate

- anonymous

okay okay one second let me do that

- anonymous

i also said differentiate.

- jim_thompson5910

Fundamental Theorem of Calculus (one of the 2 parts)
If \[\Large g(x) = \int_{a}^{x}f(t)dt\] then \[\Large g \ '(x) = f(x)\]

- jim_thompson5910

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

- anonymous

sorry, i have not used that theorem.

- anonymous

thank you for reminding me.

- anonymous

okay im getting \[\frac{-(2x+1) }{ (x^2+x+1)^2}\]

- anonymous

looks good to me

- jim_thompson5910

agreed

- anonymous

i would have written the numerator as
\[-2x-1\] so that you can solve
\[-2x-1>0\] rather easily

- anonymous

okay out of curiosity whats the second part of the theorem ... it sounds very "fundamental "

- anonymous

that is that \(F(b)-F(a)\) business

- jim_thompson5910

Other part of FTC
\[\Large \int_{a}^{b}f \ '(x) dx = f(b) - f(a)\]

- anonymous

arh damn i don't know who to give the medal too you were all so helpful

- jim_thompson5910

Most books make F ' = f, which is confusing to me because there are 2 different f's to keep track of

- anonymous

thats weird every time i see F it means \[intf(x)\]

- jim_thompson5910

F ' = f is equivalent to saying "F is the indefinite integral of f"

- anonymous

notation like to be the death of us all
nothing worse than having two identical notations for different things

- anonymous

ill keep that in mind :D
thanks again every one !!

- jim_thompson5910

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

- jim_thompson5910

no problem

- anonymous

okay real quick so i should graph that and where its positive is the integral that the original graph is concave up right?!?

- anonymous

it is concave up where
\[-2x-1>0\]

- jim_thompson5910

Exactly. The denominator is always positive, so you can ignore that portion and focus on where the numerator is positive.

- anonymous

so it would be (-infinity , -.5) right?

- anonymous

okay ill go ahead and take that as a silent nod

- jim_thompson5910

yes when x < -1/2

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