## 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?

1. Shini

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

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

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

4. 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?

5. Shini

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

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

I mean Y" oops.

8. bluebird

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

9. wio

Ummm still need help?

10. wio

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

The there is another way to find it?

12. bluebird

*then

13. wio

Well you sort of have to guess.

14. wio

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

15. wio

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

16. wio

17. bluebird

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

18. bluebird

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

19. wio

Um ok still here?

20. wio

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

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

22. 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?

23. wio

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

24. wio

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

25. bluebird

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

26. wio

why is there a negative in front of the 4?

27. bluebird

oh oops i mean 4

28. wio

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

29. wio

Did they give you y'' or y originally?

30. bluebird

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

31. wio

What was y?

32. bluebird

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

33. wio

Oh wow, what was y'?

34. wio

Why were you opposed to just graphing it?

35. bluebird

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

36. bluebird

you mean graph it with a calculator or by hand?

37. wio

by hand

38. bluebird

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

39. wio

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

or bisection

41. bluebird

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

42. wio

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

43. bluebird

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

44. wio

Seems like the only thing you could really do is to

45. wio

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

46. bluebird

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

47. wio

yeah. The problem seems kinda messed up.

48. wio

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

49. bluebird

Okay, thank you so much for your help!