\section*{Problem 3 - 101233}

Let \(f\) be the function defined for all real numbers \(x\) by \(f(x)=\frac{1-x^{2}}{x^{6}+4}\).

It is to be decided whether among all function values \(f(x)\) there is a greatest and a smallest value. These values are to be determined if they exist.
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Since \(x^{2} \geq 0\), the question is equivalent to the investigation of the function \(g(z):=\frac{1-z}{z^{3}+4}\), where only those \(z\) with \(z \geq 0\) are considered, because \(g\left(x^{2}\right)=f(x)\).

Because \(z \geq 0\), it follows that \(1-z \leq 1\) and \(z^{3}+4 \geq 4>0\), thus \(g(z) \leq \frac{1}{4}\). Indeed, \(g(0)=f(0)=\frac{1}{4}\), so this is the largest function value that \(g\) or \(f\) can take in their respective domains.

We now consider the function

\[
g^{\prime}(z)=\frac{-\left(z^{3}+4\right)-(1-z) \cdot 3 z^{2}}{\left(z^{3}+4\right)^{2}}=-\frac{1}{\left(z^{3}+4\right)^{2}} \cdot\left(z^{3}+4+3 z^{2}-3 z^{3}\right)
\]

It is

\[
g^{\prime}(z)=0 \Leftrightarrow z^{3}+4+3 z^{2}-3 z^{3}=0 \Leftrightarrow 0=2 z^{3}-3 z^{2}-4=(z-2) \cdot\left(2 z^{2}+z+2\right)
\]

Since

\[
2 z^{2}+z+2=\left(z^{2}+2 z+1\right)+z^{2}+1=(z+1)^{2}+z^{2}+1 \geq 1>0
\]

holds, \(g^{\prime}(z)\) vanishes exactly at \(z=2\).

In the interval \([0 ; \infty)\), \(g\) attains its global maximum of \(\frac{1}{4}>0\) at \(z=0\), as already seen. For \(z>1\), \(g(z)\) is even negative, but due to the higher degree of the polynomial in the denominator compared to that in the numerator, \(\lim _{z \rightarrow \infty} g(z)=0\). Therefore, \(g\) cannot be monotonically decreasing, and since it is continuously differentiable, its derivative must have at least one zero where the function \(g\) changes from monotonically decreasing to monotonically increasing. This point is uniquely determined as \(z=2\), as calculated above.

Thus, \(g(z)\) is monotonically decreasing for all \(z\) in the interval \([0,2]\) and monotonically increasing for all \(z\) in the interval \([2, \infty)\), so at \(z=2\) the function \(g\) attains its global minimum with \(g(2)=\frac{1-2}{2^{3}+4}=-\frac{1}{12}\).

Therefore, the global minimum of \(f\) is also this value \(-\frac{1}{12}\) and is attained at the points \(x= \pm \sqrt{2}\).