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?

Question

anonymous · Answer

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 · Answer

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 · Answer

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

bluebird · Answer

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 · Answer

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.