To determine the maximum value of \(\frac{x}{\sqrt{1+x^4}}\), we define the function \(f(x) = \frac{x}{\sqrt{1+x^4}}\). We will employ calculus to find its maximum.
First, we compute the derivative \(f'(x)\) using the quotient rule: If \(f(x) = \frac{u}{v}\), then \(f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2}\). Here, \(u = x\) and \(v = \sqrt{1+x^4}\).
Consequently, \(u' = 1\) and \(v' = \frac{4x^3}{2\sqrt{1+x^4}} = \frac{2x^3}{\sqrt{1+x^4}}\).
Substituting these into the quotient rule yields: \(f'(x) = \frac{\sqrt{1+x^4} \cdot 1 - x \cdot \frac{2x^3}{\sqrt{1+x^4}}}{1+x^4}\)
= \(\frac{\sqrt{1+x^4} - \frac{2x^4}{\sqrt{1+x^4}}}{1+x^4}\)
= \(\frac{(\sqrt{1+x^4})^2 - 2x^4}{(1+x^4)\sqrt{1+x^4}}\)
= \(\frac{1+x^4 - 2x^4}{(1+x^4)\sqrt{1+x^4}}\)
= \(\frac{1-x^4}{(1+x^4)\sqrt{1+x^4}}\).
Setting \(f'(x) = 0\) produces \(1-x^4 = 0\), which implies \(x^4 = 1\). The critical points are therefore \(x = \pm1\).
We evaluate \(f(x)\) at these critical points:
For \(x = 1\), \(f(1) = \frac{1}{\sqrt{1+1^4}} = \frac{1}{\sqrt{2}}\).
For \(x = -1\), \(f(-1) = \frac{-1}{\sqrt{1+(-1)^4}} = -\frac{1}{\sqrt{2}}\).
The maximum value attained is \(\frac{1}{\sqrt{2}}\).
Thus, the maximum possible value of \(\frac{x}{\sqrt{1+x^4}}\) is \(\frac{1}{\sqrt{2}}\).