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bluebird

  • one year ago

Would someone please explain how to find the inflection point, y"=0, of the equation y"= -2[4x^3-15x^2+12x-5]/(1-x^2)^3 without a graphing calculator, please?

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  1. Shini
    • one year ago
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    You already have the second derivative of the function so all you have to do is set the equation equal to 0 and then solve it. This will give you possible points of inflexion. Then you check the concavity of the point found by substituting the x-values one higher then the x-value you found and one lower. If you get a positive on one side and a negative on the other (doesn't matter which) you have found a point of inflexion.

  2. bluebird
    • one year ago
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    so is it like this? \[0=\frac{ -2\left[ 4\chi ^{3}-15\chi ^{2}+12\chi -5 \right] }{ \left( 1-\chi ^{2} \right)^{3} }\] But how do you start to solve it?

  3. Shini
    • one year ago
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    \[-2[4x^3 -15x^2 + 12x - 5] = 0\] \[4x^3 - 15x^2 + 12x -5 = 0\] Then solve it from there

  4. bluebird
    • one year ago
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    okay so I got: \[\chi \left( 4\chi ^{2}-15\chi +12 \right)=5\] then I used the \[\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\] and got 2.593070331 and 1.156929669. So do I put those numbers in the equation and test if they equal to 0?

  5. Shini
    • one year ago
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    There is also the value of x = 5. When you used the quadratic formulas to find the x-values did you first separate the equation you got into: \[x = 5, 4x^2 - 15x + 12 = 5\] If not, try that. Were you given the original equation? If you were you substitute the values you got into that to get the possible points. If not don't worry. After that i usually check if the concavity changes by drawing: |dw:1358129096454:dw| and then repeat that for the rest of the x values you got. Sorry im not being a great deal of help anymore, if anything im confusing you more.

  6. bluebird
    • one year ago
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    the original equation was \[y =\frac{ 2\chi ^{2}-4\chi +3 }{ 1-\chi^{2} }\] and I got, When X=4 Y=0.21007 When X=6 Y= 0.05133

  7. bluebird
    • one year ago
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    I mean Y" oops.

  8. bluebird
    • one year ago
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    I still can't find the inflection point...

  9. wio
    • one year ago
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    Ummm still need help?

  10. wio
    • one year ago
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    Since you have an equation of degree 3, you need to find one of the roots first without the help of the quadratic equation.

  11. bluebird
    • one year ago
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    The there is another way to find it?

  12. bluebird
    • one year ago
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    *then

  13. wio
    • one year ago
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    Well you sort of have to guess.

  14. wio
    • one year ago
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    What is the equation we want to find the roots of exactly?

  15. wio
    • one year ago
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    Which function exactly is \(y''\). I see a lot of functions here.

  16. wio
    • one year ago
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    @bluebird Please?

  17. bluebird
    • one year ago
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    sorry about that... the first one equation in the question is y". I need to find the inflection point for y"=0

  18. bluebird
    • one year ago
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    \[y"=\frac{ -2\left[ 4\chi ^{3}-15\chi ^{2}+12\chi -5 \right] }{ \left( 1-\chi ^{2} \right)^{3} }\] to be exact

  19. wio
    • one year ago
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    Um ok still here?

  20. wio
    • one year ago
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    First thing to notice is that it's undefined at 1 and -1. Next thing to notice is that if the numerator (part on top) is 0, then the whole thing is 0. So we can ignore the denominator as long as we keep the first thing in mind.

  21. wio
    • one year ago
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    I'd start by just distributing that \(-2\). There's not reason to have that factored out.

  22. bluebird
    • one year ago
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    oh okay... so it's only the \[-2\left[ 4\chi ^{3}-15\chi ^{2}+12\chi -5 \right]\] part I work with right?

  23. wio
    • one year ago
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    Yes. Ultimately we could multiply both sides by the denominator, and since the other size is 0, it would remain 0.

  24. wio
    • one year ago
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    I suppose you could get rid of that \(-2\) with the same logic.

  25. bluebird
    • one year ago
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    \[-4\chi^{3}-15\chi ^{2}+12\chi -5\] is not factor-able right?

  26. wio
    • one year ago
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    why is there a negative in front of the 4?

  27. bluebird
    • one year ago
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    oh oops i mean 4

  28. wio
    • one year ago
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    yeah apparently it has a really nasty real root, and two imaginary roots.

  29. wio
    • one year ago
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    Did they give you y'' or y originally?

  30. bluebird
    • one year ago
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    they gave me y originally and I was suppose to sketch the graph and find the inflection points without a graphing calculator

  31. wio
    • one year ago
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    What was y?

  32. bluebird
    • one year ago
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    \[y=\frac{ 2\chi ^{2}-4\chi +3 }{1-\chi ^{2} }\]

  33. wio
    • one year ago
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    Oh wow, what was y'?

  34. wio
    • one year ago
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    Why were you opposed to just graphing it?

  35. bluebird
    • one year ago
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    \[y'=\frac{ -4\chi ^{2}+10\chi -4 }{ \left( 1-\chi ^{2} \right)^{2} }\]

  36. bluebird
    • one year ago
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    you mean graph it with a calculator or by hand?

  37. wio
    • one year ago
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    by hand

  38. bluebird
    • one year ago
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    I have to label the inflection points and I can't find the y"=0 one

  39. wio
    • one year ago
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    The only way to really find it is to use the very erudite formula for 3rd degree polynomials, or use some other root finding method.... like Newton's method.

  40. wio
    • one year ago
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    or bisection

  41. bluebird
    • one year ago
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    I don't know the other methods....

  42. wio
    • one year ago
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    http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_roots The formula is HUGE

  43. bluebird
    • one year ago
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    I'm not sure if i can memorize that....

  44. wio
    • one year ago
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    Seems like the only thing you could really do is to

  45. wio
    • one year ago
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    graph it and then plug in points really close to the inflection point

  46. bluebird
    • one year ago
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    oh okay... so I can just guess the inflection point then, right?

  47. wio
    • one year ago
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    yeah. The problem seems kinda messed up.

  48. wio
    • one year ago
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    You can sort of see how the tangent line would intersect between 2 and 3

  49. bluebird
    • one year ago
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    Okay, thank you so much for your help!

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