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