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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Ashleyisakitty @ganeshie8 @Hero @jim_thompson5910

  3. anonymous
    • one year ago
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    i know i have to find the second derivative im just not sure how

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

  5. anonymous
    • one year ago
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    wow okay give me a second to read that...

  6. jim_thompson5910
    • one year ago
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    @surjithayer there's no need to integrate you just need to find \(\Large \frac{d^2y}{dx^2}\)

  7. anonymous
    • one year ago
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    oh no first derivative is the integrand

  8. anonymous
    • one year ago
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    oh what @jim_thompson5910 said

  9. jim_thompson5910
    • one year ago
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    the first derivative is not too bad because you can use the fundamental theorem of calculus

  10. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    interval is 0 to x or we can change it to x then diff.

  13. anonymous
    • one year ago
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    i mean derive not integrate

  14. anonymous
    • one year ago
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    yeah plug in x for t

  15. anonymous
    • one year ago
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    then DIFFERENTIATE that , not integrate

  16. anonymous
    • one year ago
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    okay okay one second let me do that

  17. anonymous
    • one year ago
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    i also said differentiate.

  18. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    sorry, i have not used that theorem.

  21. anonymous
    • one year ago
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    thank you for reminding me.

  22. anonymous
    • one year ago
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    okay im getting \[\frac{-(2x+1) }{ (x^2+x+1)^2}\]

  23. anonymous
    • one year ago
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    looks good to me

  24. jim_thompson5910
    • one year ago
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    agreed

  25. anonymous
    • one year ago
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    i would have written the numerator as \[-2x-1\] so that you can solve \[-2x-1>0\] rather easily

  26. anonymous
    • one year ago
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    okay out of curiosity whats the second part of the theorem ... it sounds very "fundamental "

  27. anonymous
    • one year ago
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    that is that \(F(b)-F(a)\) business

  28. jim_thompson5910
    • one year ago
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    Other part of FTC \[\Large \int_{a}^{b}f \ '(x) dx = f(b) - f(a)\]

  29. anonymous
    • one year ago
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    arh damn i don't know who to give the medal too you were all so helpful

  30. jim_thompson5910
    • one year ago
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    Most books make F ' = f, which is confusing to me because there are 2 different f's to keep track of

  31. anonymous
    • one year ago
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    thats weird every time i see F it means \[intf(x)\]

  32. jim_thompson5910
    • one year ago
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    F ' = f is equivalent to saying "F is the indefinite integral of f"

  33. anonymous
    • one year ago
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    notation like to be the death of us all nothing worse than having two identical notations for different things

  34. anonymous
    • one year ago
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    ill keep that in mind :D thanks again every one !!

  35. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    no problem

  37. anonymous
    • one year ago
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    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
    • one year ago
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    it is concave up where \[-2x-1>0\]

  39. jim_thompson5910
    • one year ago
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    Exactly. The denominator is always positive, so you can ignore that portion and focus on where the numerator is positive.

  40. anonymous
    • one year ago
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    so it would be (-infinity , -.5) right?

  41. anonymous
    • one year ago
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    okay ill go ahead and take that as a silent nod

  42. jim_thompson5910
    • one year ago
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    yes when x < -1/2

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