\section*{Exercise 3 - 061233}

All real numbers \(x\) in the intervals \(0<x<\frac{\pi}{2}\) and \(\frac{\pi}{2}<x<\pi\) are to be specified for which

\[
f(x)=\sin x+\cos x+\tan x+\cot x
\]

is positive, and all real numbers \(x\), in the same intervals, for which \(f(x)\) is negative. Is there a smallest positive value that \(f(x)\) assumes in the above intervals, and if so, what is this value?
}

For \(0 < x < \frac{\pi}{2}\), \(\cos x > 0\) and \(\cos x > -1\). Therefore, \(\frac{1}{\cos x} > 0\). Since \(\tan^2 x \geq 0\), we have \(f_{2}(x) = \tan^2 x + \frac{1}{\cos x} \geq 2\), with equality holding only for \(\tan x = 1\), i.e., \(x = \frac{\pi}{4}\) as the only value in the considered interval.

For the analysis of \(f_{1}\), we consider its derivative \(f_{1}^{\prime}(x) = \cos x - \sin x\), which in the considered interval vanishes only for \(x = \frac{\pi}{4}\). Since the cosine function is strictly decreasing and the sine function is strictly increasing in this interval, \(f_{1}^{\prime}\) is strictly decreasing and thus takes positive values for arguments \(x\) smaller than \(\frac{\pi}{4}\) and negative values for larger arguments.

Therefore, the function \(f_{1}\) is strictly increasing in the interval \(0 < x < \frac{\pi}{4}\) and strictly decreasing in the interval \(\frac{\pi}{4} < x < \frac{\pi}{2}\). Thus,

\[
f_{1}(x) \geq f_{1}\left(\frac{\pi}{4}\right) = \sin \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}
\]

with equality holding only for \(x = \frac{\pi}{4}\).

Together, this yields

\[
f(x) = f_{1}(x) + f_{2}(x) \geq \sqrt{2} + 2
\]

with equality holding exactly for \(x = \frac{\pi}{4}\). Therefore, \(2 + \sqrt{2}\) is the smallest positive value that \(f\) attains in the considered interval.