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

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

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.

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"

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

no problem

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

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

okay ill go ahead and take that as a silent nod

yes when x < -1/2