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AravindG

Find the points of maxima/minima of

  • one year ago
  • one year ago

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  1. hal_stirrup
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    whats you Q?

    • one year ago
  2. AravindG
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    \[\int\limits^{x^2}_0 [(t^2-5t+4)/(2+e^t)]dt\]

    • one year ago
  3. AravindG
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    @amistre64 , @UnkleRhaukus , @Callisto , @Ishaan94

    • one year ago
  4. AravindG
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    @.Sam.

    • one year ago
  5. AravindG
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    @cwrw238

    • one year ago
  6. UnkleRhaukus
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    (t2−5t+4)=(t-1)(t-4)

    • one year ago
  7. hal_stirrup
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    am not sure if you need to know the answer or how to go about answering the question.

    • one year ago
  8. AravindG
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    ?

    • one year ago
  9. AravindG
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    help pls

    • one year ago
  10. hal_stirrup
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    so i can help. i need to know what are you really looking for?

    • one year ago
  11. AravindG
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    maxima and minima !!

    • one year ago
  12. cwrw238
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    the expression in the integral = 0 for max/minm points ( i think) isn't that right guys?

    • one year ago
  13. Spacelimbus
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    I would have guessed so @cwrw238, I am not sure if the integrated function and the to be integrated function share similarities in minima and maxima though. but for linear functions that seems to be the case (-:

    • one year ago
  14. cwrw238
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    so (t - 4)(t - 1) = 0 t = 4 or 1 so you have turning points at values t = 4 and t = 1 second derivative is 2t - 5 when t = 4 this is positive so t=4 is a minimum when t = 1 this is negative so t = 1 is a maximum

    • one year ago
  15. cwrw238
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    yea spacelimbus - im not sure either!

    • one year ago
  16. hal_stirrup
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    integral (4-5 t+t^2)/(2+e^t) dt = (5/2-t) Li_2(-e^t/2)+Li_3(-e^t/2)+t^3/6-(5 t^2)/4-1/2 t^2 log(1/2 (e^t+2))+2 t+5/2 t log(1/2 (e^t+2))-2 log(e^t+2)+constant

    • one year ago
  17. Spacelimbus
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    THe polylogarithmic function behaves weirdly for min/max though @hal_stirrup , I wouldn't know how to solve that expression for zeros.

    • one year ago
  18. Spacelimbus
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    @cwrw238, lets think about this together, an integral is the antiderivative of a function right? And to find minima and maxima you need to set the derivative to zero, correct?

    • one year ago
  19. hal_stirrup
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    yes

    • one year ago
  20. Spacelimbus
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    therefore, what we have next to the integral expression is the derivative of the function, I would say your answer is pretty much correct.

    • one year ago
  21. Spacelimbus
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    is the derivative of the integrated function*

    • one year ago
  22. cwrw238
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    makes sense to me -though i'm learning calculus at present time so I'm not expert by any means

    • one year ago
  23. cwrw238
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    thats some integral!!!! lol

    • one year ago
  24. hal_stirrup
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    ok integral is the area under the curve.

    • one year ago
  25. cwrw238
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    yes

    • one year ago
  26. Spacelimbus
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    Oh okay @cwrw238, I believe it's correct, it's how I would read it anyway, it's basically an easier way (that looks more complicated at first glance though) If you integrate an arbitary function you get F(x), if you derive that again you get f(x), the derivative of a function is what decides if there are minima and/or maxima points present. So in our case, just the f(x) =0, because that is the derivative of the obscure function I wouldn't know how to integrate without series and transformations *grins*

    • one year ago
  27. cwrw238
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    nor me!! - yes i think these values of t correspond to max ./ mini on the graph . just noticed the limits on integral - i expect that t^2 not x^2

    • one year ago
  28. Spacelimbus
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    Seems to be a time -> position transformation, but I cannot tell, however, the points we have found are extremal points, but to decide wether they are min/max or not we need the second derivative of the function f(x) which is a bit more complicated since it requires the use of the quotient rule.

    • one year ago
  29. hal_stirrup
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    typo error cwrw238 i think too.

    • one year ago
  30. cwrw238
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    oh yes of course - i only differentiated the numerator!!! my mistake there.

    • one year ago
  31. cwrw238
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    typo?

    • one year ago
  32. Spacelimbus
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    not a big thing, just some additional work (-: but @AravindG can verify these points for himself.

    • one year ago
  33. cwrw238
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    i just entered the integral on wolfram alpha http://www.wolframalpha.com/input/?i=integrate+++%28t%5E2−5t%2B4%29%2F%282%2Be%5Et%29+ - theres no min/max on the plot at the points t = 1 and 4 so it looks like i'm wrong...

    • one year ago
  34. hal_stirrup
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    ok i recommend 1: to go back the question and se if there is a typo error. 2: go back to basic integration methods in definite integral . and you will be able to solve it .

    • one year ago
  35. hal_stirrup
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    example :\[\int\limits_{}^{} x^4\] its primative funaction is : \[x ^{5}/5\]

    • one year ago
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