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

1. anonymous

2. anonymous

@Ashleyisakitty @ganeshie8 @Hero @jim_thompson5910

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. jim_thompson5910

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

7. anonymous

oh no first derivative is the integrand

8. anonymous

oh what @jim_thompson5910 said

9. jim_thompson5910

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

10. 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?

11. 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

12. anonymous

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

13. anonymous

i mean derive not integrate

14. anonymous

yeah plug in x for t

15. anonymous

then DIFFERENTIATE that , not integrate

16. anonymous

okay okay one second let me do that

17. anonymous

i also said differentiate.

18. jim_thompson5910

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

19. 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

20. anonymous

sorry, i have not used that theorem.

21. anonymous

thank you for reminding me.

22. anonymous

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

23. anonymous

looks good to me

24. jim_thompson5910

agreed

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. jim_thompson5910

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

29. anonymous

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

30. jim_thompson5910

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

31. anonymous

thats weird every time i see F it means $intf(x)$

32. jim_thompson5910

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

33. anonymous

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

34. anonymous

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

35. 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

36. jim_thompson5910

no problem

37. anonymous

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

38. anonymous

it is concave up where $-2x-1>0$

39. jim_thompson5910

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

40. anonymous

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

41. anonymous

okay ill go ahead and take that as a silent nod

42. jim_thompson5910

yes when x < -1/2