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bluebird

  • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    I mean Y" oops.

  8. bluebird
    • 2 years ago
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    I still can't find the inflection point...

  9. wio
    • 2 years ago
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    Ummm still need help?

  10. wio
    • 2 years 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
    • 2 years ago
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    The there is another way to find it?

  12. bluebird
    • 2 years ago
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    *then

  13. wio
    • 2 years ago
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    Well you sort of have to guess.

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

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

  16. wio
    • 2 years ago
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    @bluebird Please?

  17. bluebird
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Um ok still here?

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

  22. bluebird
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    I suppose you could get rid of that \(-2\) with the same logic.

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

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

  27. bluebird
    • 2 years ago
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    oh oops i mean 4

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

  29. wio
    • 2 years ago
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    Did they give you y'' or y originally?

  30. bluebird
    • 2 years 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
    • 2 years ago
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    What was y?

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

  33. wio
    • 2 years ago
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    Oh wow, what was y'?

  34. wio
    • 2 years ago
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    Why were you opposed to just graphing it?

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

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

  37. wio
    • 2 years ago
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    by hand

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

  39. wio
    • 2 years 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
    • 2 years ago
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    or bisection

  41. bluebird
    • 2 years ago
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    I don't know the other methods....

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

  43. bluebird
    • 2 years ago
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    I'm not sure if i can memorize that....

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

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

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

  47. wio
    • 2 years ago
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    yeah. The problem seems kinda messed up.

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

  49. bluebird
    • 2 years ago
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    Okay, thank you so much for your help!

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