## bluebird 3 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?

1. 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.

2. anonymous

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. anonymous

$-2[4x^3 -15x^2 + 12x - 5] = 0$ $4x^3 - 15x^2 + 12x -5 = 0$ Then solve it from there

4. anonymous

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. 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.

6. anonymous

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. anonymous

I mean Y" oops.

8. anonymous

I still can't find the inflection point...

9. anonymous

Ummm still need help?

10. 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.

11. anonymous

The there is another way to find it?

12. anonymous

*then

13. anonymous

Well you sort of have to guess.

14. anonymous

What is the equation we want to find the roots of exactly?

15. anonymous

Which function exactly is $$y''$$. I see a lot of functions here.

16. anonymous

17. anonymous

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

18. anonymous

$y"=\frac{ -2\left[ 4\chi ^{3}-15\chi ^{2}+12\chi -5 \right] }{ \left( 1-\chi ^{2} \right)^{3} }$ to be exact

19. anonymous

Um ok still here?

20. 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.

21. anonymous

I'd start by just distributing that $$-2$$. There's not reason to have that factored out.

22. anonymous

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. anonymous

Yes. Ultimately we could multiply both sides by the denominator, and since the other size is 0, it would remain 0.

24. anonymous

I suppose you could get rid of that $$-2$$ with the same logic.

25. anonymous

$-4\chi^{3}-15\chi ^{2}+12\chi -5$ is not factor-able right?

26. anonymous

why is there a negative in front of the 4?

27. anonymous

oh oops i mean 4

28. anonymous

yeah apparently it has a really nasty real root, and two imaginary roots.

29. anonymous

Did they give you y'' or y originally?

30. anonymous

they gave me y originally and I was suppose to sketch the graph and find the inflection points without a graphing calculator

31. anonymous

What was y?

32. anonymous

$y=\frac{ 2\chi ^{2}-4\chi +3 }{1-\chi ^{2} }$

33. anonymous

Oh wow, what was y'?

34. anonymous

Why were you opposed to just graphing it?

35. anonymous

$y'=\frac{ -4\chi ^{2}+10\chi -4 }{ \left( 1-\chi ^{2} \right)^{2} }$

36. anonymous

you mean graph it with a calculator or by hand?

37. anonymous

by hand

38. anonymous

I have to label the inflection points and I can't find the y"=0 one

39. 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.

40. anonymous

or bisection

41. anonymous

I don't know the other methods....

42. anonymous

http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_roots The formula is HUGE

43. anonymous

I'm not sure if i can memorize that....

44. anonymous

Seems like the only thing you could really do is to

45. anonymous

graph it and then plug in points really close to the inflection point

46. anonymous

oh okay... so I can just guess the inflection point then, right?

47. anonymous

yeah. The problem seems kinda messed up.

48. anonymous

You can sort of see how the tangent line would intersect between 2 and 3

49. anonymous

Okay, thank you so much for your help!