\section*{Exercise 2 - 171012}

Determine the set of all real numbers \(x\) for which the term \(\frac{1}{\sqrt{33-8 x-x^{2}}}\) is defined!
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The term \(\sqrt{33-8 x-x^{2}}\) is defined if and only if \(33-8 x-x^{2} \geq 0\).

The term \(\frac{1}{\sqrt{33-8 x-x^{2}}}\) is defined if and only if \(\sqrt{33-8 x-x^{2}}\) is defined and \(\sqrt{33-8 x-x^{2}} \neq 0\). These requirements are successively equivalent to

\[
\begin{array}{cccccc}
33-8 x-x^{2}>0 & \rightarrow & x^{2}+8 x+16<49 & \rightarrow & (x-4)^{2}<49 \quad \rightarrow & |x-4|<7 \\
0 \leq x+4<7 & \text { or } & -7<x+4<0 \quad \rightarrow \quad-4 \leq x<3 \quad \text { or } & -11<x<-4
\end{array}
\]

Therefore, the set of all \(x\) for which \(-11<x<3\) is the desired set.