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bluebird
 4 years ago
Would someone please explain how to find the inflection point, y"=0, of the equation y"= 2[4x^315x^2+12x5]/(1x^2)^3 without a graphing calculator, please?
bluebird
 4 years ago
Would someone please explain how to find the inflection point, y"=0, of the equation y"= 2[4x^315x^2+12x5]/(1x^2)^3 without a graphing calculator, please?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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 xvalues one higher then the xvalue 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
 4 years ago
Best ResponseYou've already chosen the best response.0so 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[2[4x^3 15x^2 + 12x  5] = 0\] \[4x^3  15x^2 + 12x 5 = 0\] Then solve it from there

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0okay 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
 4 years ago
Best ResponseYou've already chosen the best response.0There is also the value of x = 5. When you used the quadratic formulas to find the xvalues 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
 4 years ago
Best ResponseYou've already chosen the best response.0the 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
 4 years ago
Best ResponseYou've already chosen the best response.0I still can't find the inflection point...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ummm still need help?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since you have an equation of degree 3, you need to find one of the roots first without the help of the quadratic equation.

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0The there is another way to find it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well you sort of have to guess.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What is the equation we want to find the roots of exactly?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Which function exactly is \(y''\). I see a lot of functions here.

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0sorry about that... the first one equation in the question is y". I need to find the inflection point for y"=0

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0\[y"=\frac{ 2\left[ 4\chi ^{3}15\chi ^{2}+12\chi 5 \right] }{ \left( 1\chi ^{2} \right)^{3} }\] to be exact

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First 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
 4 years ago
Best ResponseYou've already chosen the best response.0I'd start by just distributing that \(2\). There's not reason to have that factored out.

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0oh okay... so it's only the \[2\left[ 4\chi ^{3}15\chi ^{2}+12\chi 5 \right]\] part I work with right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. Ultimately we could multiply both sides by the denominator, and since the other size is 0, it would remain 0.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I suppose you could get rid of that \(2\) with the same logic.

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0\[4\chi^{3}15\chi ^{2}+12\chi 5\] is not factorable right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why is there a negative in front of the 4?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah apparently it has a really nasty real root, and two imaginary roots.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Did they give you y'' or y originally?

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0they gave me y originally and I was suppose to sketch the graph and find the inflection points without a graphing calculator

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0\[y=\frac{ 2\chi ^{2}4\chi +3 }{1\chi ^{2} }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why were you opposed to just graphing it?

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0\[y'=\frac{ 4\chi ^{2}+10\chi 4 }{ \left( 1\chi ^{2} \right)^{2} }\]

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0you mean graph it with a calculator or by hand?

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0I have to label the inflection points and I can't find the y"=0 one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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.

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know the other methods....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_roots The formula is HUGE

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not sure if i can memorize that....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Seems like the only thing you could really do is to

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0graph it and then plug in points really close to the inflection point

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0oh okay... so I can just guess the inflection point then, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah. The problem seems kinda messed up.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You can sort of see how the tangent line would intersect between 2 and 3

bluebird
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, thank you so much for your help!
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