\section*{Exercise 2 - 231012}

Determine all real numbers \(x\) for which \(x^{3}-9 x>0\) holds!
}

It holds that

\[
x^{3}-9 x=x\left(x^{2}-9\right)=x(x-3)(x+3)
\]

Thus, the left side of the inequality can be represented as a product of three factors. A product of three factors is positive if and only if

(a) all three factors are positive or

(b) two factors are negative and the third one is positive.

Case (b) occurs exactly when the second largest factor \(x\) is negative and the largest factor \(x+3\) is positive, i.e., exactly when the inequalities \(x-3\) hold, and case (a) exactly when \(x-3\) is positive, i.e., \(x>3\).

Therefore, the real numbers \(x\) to be determined are exactly those for which:

\[
-3 < x < 0 \quad \text{or} \quad x > 3
\]