1. At which x-value(s) does f(x) = (x^3)/ (1+x^4)
attain its absolute maximum value?

I took the first derivative and ended up getting x=0 and 3^(1/4). Are they correct and how do I find out which one is min or max, ect.?

Question

1. At which x-value(s) does f(x) = (x^3)/ (1+x^4)
attain its absolute maximum value?

I took the first derivative and ended up getting x=0 and 3^(1/4). Are they correct and how do I find out which one is min or max, ect.?

anonymous · Answer

Assuming those are correct, you would plug them into the originl function and see which one gave you the largest value. The original function is always going to give you placement, like min, max, where it is, etc.

anonymous · Answer

So you get the derivative and end up setting $3x^{2}-x^{6}=0$. Solving for this you get $x^{2}(3-x^{4})$= 0
so x = 0 
$x^{4} = 3\ \implies\ x = \pm 3^{1/4}$
So looks like you just missed a -3^(1/4), otherwise its correct what you have.

anonymous · Answer

OKay, thanks! 3^ (1/4) is such an awkward number :(

anonymous · Answer

Eh, it happens. Not like you have to know what it is, you just accept it and do what you need with it, lol.