I'm having trouble finding the Points of Inflection for this function

[agv567]

Homework Statement

It asks for any Points of Inflection
The second derivative is given: y'' = (2x^5 + 4x^3 - 6x) / (x^2-1)^4

Homework Equations

The original function was x / (X^2 -1)

The Attempt at a Solution

I have set it equal to zero, and I get 2x^5 + 4x^3 - 6x = 0

I then simplified, taking out the 2, as well as factoring out an x

leaving me with

x(x^2 + 3)(x^2 -1)...

I get a weird answer though...x^2=-3 does not exist...and x = =+-1, are vertical asymptotes in the original function

help?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[agv567]

help...anyone ;_;

 

[SammyS]

help...anyone ;_;

Hi agv567. Welcome to PF.

Sometimes it takes a while for those of us who volunteer on the site to get around to answering questions. That's why there is a rule which forbids "bumping" your post until it has been here 24 hours.

I'll look at your question & get back to you soon.

 

[eumyang]

Homework Statement

It asks for any Points of Inflection
The second derivative is given: y'' = 2x^5 + 4x^2 - 6x / (x^2-1)^4

This can be simplified further. Also, are you sure the numerator isn't
2x⁵ + 4x³ - 6x?

I then simplified, taking out the 2, as well as factoring out an x

leaving me with

x(x^2 + 3)(x^2 -1)...

I get a weird answer though...x^2=-3 does not exist...and x = =+-1, are vertical asymptotes in the original function

There's one more factor to check, the one in bold above.

 

[agv567]

EDIT: I'm an idiot

And POI can't occur at asymptotes, right? so -1 or 1 are not POI?

and x = 0...so the only POI occurs at (0,0), right? I checked with my calculator and it seems so.But what about the x^2 = -3 equation? Since it does not exist...I do nothing with it?

 

[SammyS]

Homework Statement

It asks for any Points of Inflection
The second derivative is given: y'' = 2x^5 + 4x^2 - 6x / (x^2-1)^4

Homework Equations

The original function was x / (X^2 -1)

The Attempt at a Solution

I have set it equal to zero, and I get 2x^5 + 4x^2 - 6x = 0

I then simplified, taking out the 2, as well as factoring out an x

leaving me with

x(x^2 + 3)(x^2 -1)...

I get a weird answer though...x^2=-3 does not exist...and x = =+-1, are vertical asymptotes in the original function

help?

You have a typo in that second derivative. It's y'' = (2x⁵ + 4x³ - 6x) / (x^2-1)^4

That should fix-up your problem!

Also, please use proper parentheses. Otherwise what you would have is equivalent to: $\displaystyle 2x^5 + 4x^3 -\frac{6x}{(x^2-1)^4}$

 

[agv567]

Thank you so much. I see my mistake now and how it confused everyone haha. I still don't understand what to do with the x^2 = -3, though.

 

[SammyS]

x²=-3 doesn't have any real number solutions !