anonymous
  • anonymous
Find the interval on which the curve of "picture below" is concave up
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
@Ashleyisakitty @ganeshie8 @Hero @jim_thompson5910
anonymous
  • anonymous
i know i have to find the second derivative im just not sure how

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.