Confirm my derivative for f(x) = (x^2 - 1)^2 /(x^4 + 1)

[hank]

f(x) = (x^2 - 1)^2 /(x^4 + 1)

f'(x) = 4x(x^4 - 1) //Top divided by bottom.
--------------
(x^4 + 1)^2


Does this look right?

 

[galactus]

Looks good.

 

[hank]

Ok, cool...

So is f''(x), the second derivative, (-12x^8 + 48x^4) / (x^4 + 1) ?

Thanks!

 

[galactus]

Actually, the 2nd derivative is

\(\displaystyle \L\\\frac{-4(3x^{8}-12x^{4}+1)}{(x^{4}+1)^{3}}\)

Try it again.

 

[hank]

Ok, I got it this time.

My question now, is how would I find x if I set the whole thing to zero?

I think I would start off by multiplying both sides by (x^4 + 1)^3.

That would leave me with:

-4(3x^8 - 12x^4 + 1) = 0

//I divide both sides by -4.
(3x^8 - 12x^4 + 1) = 0

I don't think that factors out evenly. Is it ok to use the quadratic formula, even tho it's an 8th degree polynomial?

If I do that, I end up with:

2 +/- sqrt(33)
----------------
3

Does this look right?

 

[hank]

bump