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AravindG

  • 2 years ago

Find the points of maxima/minima of

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  1. hal_stirrup
    • 2 years ago
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    whats you Q?

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

  3. AravindG
    • 2 years ago
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    @amistre64 , @UnkleRhaukus , @Callisto , @Ishaan94

  4. AravindG
    • 2 years ago
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    @.Sam.

  5. AravindG
    • 2 years ago
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    @cwrw238

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

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

  8. AravindG
    • 2 years ago
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    ?

  9. AravindG
    • 2 years ago
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    help pls

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

  11. AravindG
    • 2 years ago
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    maxima and minima !!

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

  13. Spacelimbus
    • 2 years ago
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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 (-:

  14. cwrw238
    • 2 years ago
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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

  15. cwrw238
    • 2 years ago
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    yea spacelimbus - im not sure either!

  16. hal_stirrup
    • 2 years ago
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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

  17. Spacelimbus
    • 2 years ago
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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.

  18. Spacelimbus
    • 2 years ago
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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?

  19. hal_stirrup
    • 2 years ago
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    yes

  20. Spacelimbus
    • 2 years ago
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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.

  21. Spacelimbus
    • 2 years ago
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    is the derivative of the integrated function*

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

  23. cwrw238
    • 2 years ago
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    thats some integral!!!! lol

  24. hal_stirrup
    • 2 years ago
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    ok integral is the area under the curve.

  25. cwrw238
    • 2 years ago
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    yes

  26. Spacelimbus
    • 2 years ago
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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*

  27. cwrw238
    • 2 years ago
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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

  28. Spacelimbus
    • 2 years ago
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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.

  29. hal_stirrup
    • 2 years ago
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    typo error cwrw238 i think too.

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

  31. cwrw238
    • 2 years ago
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    typo?

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

  33. cwrw238
    • 2 years ago
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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...

  34. hal_stirrup
    • 2 years ago
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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 .

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

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