bluebird
  • bluebird
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?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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.
bluebird
  • bluebird
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?
anonymous
  • anonymous
\[-2[4x^3 -15x^2 + 12x - 5] = 0\] \[4x^3 - 15x^2 + 12x -5 = 0\] Then solve it from there

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bluebird
  • bluebird
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?
anonymous
  • anonymous
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.
bluebird
  • bluebird
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
bluebird
  • bluebird
I mean Y" oops.
bluebird
  • bluebird
I still can't find the inflection point...
anonymous
  • anonymous
Ummm still need help?
anonymous
  • anonymous
Since you have an equation of degree 3, you need to find one of the roots first without the help of the quadratic equation.
bluebird
  • bluebird
The there is another way to find it?
bluebird
  • bluebird
*then
anonymous
  • anonymous
Well you sort of have to guess.
anonymous
  • anonymous
What is the equation we want to find the roots of exactly?
anonymous
  • anonymous
Which function exactly is \(y''\). I see a lot of functions here.
anonymous
  • anonymous
@bluebird Please?
bluebird
  • bluebird
sorry about that... the first one equation in the question is y". I need to find the inflection point for y"=0
bluebird
  • bluebird
\[y"=\frac{ -2\left[ 4\chi ^{3}-15\chi ^{2}+12\chi -5 \right] }{ \left( 1-\chi ^{2} \right)^{3} }\] to be exact
anonymous
  • anonymous
Um ok still here?
anonymous
  • anonymous
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.
anonymous
  • anonymous
I'd start by just distributing that \(-2\). There's not reason to have that factored out.
bluebird
  • bluebird
oh okay... so it's only the \[-2\left[ 4\chi ^{3}-15\chi ^{2}+12\chi -5 \right]\] part I work with right?
anonymous
  • anonymous
Yes. Ultimately we could multiply both sides by the denominator, and since the other size is 0, it would remain 0.
anonymous
  • anonymous
I suppose you could get rid of that \(-2\) with the same logic.
bluebird
  • bluebird
\[-4\chi^{3}-15\chi ^{2}+12\chi -5\] is not factor-able right?
anonymous
  • anonymous
why is there a negative in front of the 4?
bluebird
  • bluebird
oh oops i mean 4
anonymous
  • anonymous
yeah apparently it has a really nasty real root, and two imaginary roots.
anonymous
  • anonymous
Did they give you y'' or y originally?
bluebird
  • bluebird
they gave me y originally and I was suppose to sketch the graph and find the inflection points without a graphing calculator
anonymous
  • anonymous
What was y?
bluebird
  • bluebird
\[y=\frac{ 2\chi ^{2}-4\chi +3 }{1-\chi ^{2} }\]
anonymous
  • anonymous
Oh wow, what was y'?
anonymous
  • anonymous
Why were you opposed to just graphing it?
bluebird
  • bluebird
\[y'=\frac{ -4\chi ^{2}+10\chi -4 }{ \left( 1-\chi ^{2} \right)^{2} }\]
bluebird
  • bluebird
you mean graph it with a calculator or by hand?
anonymous
  • anonymous
by hand
bluebird
  • bluebird
I have to label the inflection points and I can't find the y"=0 one
anonymous
  • anonymous
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.
anonymous
  • anonymous
or bisection
bluebird
  • bluebird
I don't know the other methods....
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_roots The formula is HUGE
bluebird
  • bluebird
I'm not sure if i can memorize that....
anonymous
  • anonymous
Seems like the only thing you could really do is to
anonymous
  • anonymous
graph it and then plug in points really close to the inflection point
bluebird
  • bluebird
oh okay... so I can just guess the inflection point then, right?
anonymous
  • anonymous
yeah. The problem seems kinda messed up.
anonymous
  • anonymous
You can sort of see how the tangent line would intersect between 2 and 3
bluebird
  • bluebird
Okay, thank you so much for your help!

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