f(x) = x + (128) / (x^2+16). Identify all relative extrema.

Here is my progress:

After Deriving etc, I came up with:

(x^2 + 16)^2 - 256x = 0
I tried to expand and got

x^4 + 16x^4 + 256 - 256x = 0

Can someone help me solve this ?

Thanks!

Question

f(x) = x + (128) / (x^2+16). Identify all relative extrema.

Here is my progress:

After Deriving etc, I came up with:

(x^2 + 16)^2 - 256x = 0
I tried to expand and got

x^4 + 16x^4 + 256 - 256x = 0

Can someone help me solve this ?

Thanks!

abb0t · Answer

$$f' = -\frac{ x^2+256x-16 }{ (x^2+16)^2 }$$

abb0t · Answer

To find the max, you set it equal to zero. Which should make the problem fairly easy to solve.

anonymous · Answer

Can you give me some assistance with this part, I always have a hard time :(

abb0t · Answer

HINT: .
$$\frac{ ax }{ c } = 0 $$
$$c \times 0 = 0$$
hence $$ax =0$$

anonymous · Answer

Honestly I have no idea how to even solve it. :(. Can you solve it for me please?

abb0t · Answer

I already gave you the derivative. When you multiply the denominator by zero, you only get a quadratic! You don't know how to factor the quadratic? Is that where you're struggling?!

abb0t · Answer

Factor: $$-(x^2+256x-16) = 0$$

anonymous · Answer

Oh...I think I am starting to understand lol.

abb0t · Answer

$$(x \pm ?_1 ) (x \pm ?_2)$$

anonymous · Answer

So I should do what after " −(x2+256x−16)=0"

abb0t · Answer

factor it out! 
(x + ___ )( x + ___) 
solve for x.

anonymous · Answer

Uh..How do I factor that? my results dont make sense..

anonymous · Answer

what did u get >.>