Solution: Given that $f,g: \mathbb{R} \to \mathbb{R} $ be functions.
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We will prove that given any two real numbers $a, b$ we have
\begin{equation}\label{eq:17Feb2025-1}
\min \{ a,b \} = \frac{1}{2}(a + b) - \frac{1}{2}\vert a - b \vert .
\end{equation}
We will divide the proof in two cases. At first let us assume that $a \geq b$. Then $\vert a - b \vert = a - b$. Thus we have
\begin{align*}
\frac{1}{2}(a + b) - \frac{1}{2}\vert a - b \vert & = \frac{1}{2}(a + b) - \frac{1}{2}(a - b) \\
& = b = \min \{ a,b \} .
\end{align*}
On the other hand, if we assume that $a < b$, then $\vert a - b \vert = -(a - b) = b - a$. Thus,
\begin{align*}
\frac{1}{2}(a + b) - \frac{1}{2}\vert a - b \vert & = \frac{1}{2}(a + b) - \frac{1}{2}(b - a) \\
& = a = \min \{ a, b \} .
\end{align*}
Therefore, \eqref{eq:17Feb2025-1} is proved. Thus for any $x \in \mathbb{R} $,
\[
\min \{ f(x), g(x) \} = \frac{1}{2}(f(x) + g(x)) - \frac{1}{2}\vert f(x) = g(x) \vert,
\]
which implies the required claim for the functions $f$ and $g$.
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By the similar argument as in part (a), we can show that
\begin{equation}\label{eq:17Feb2025-2}
\max \{ a,b \} = \frac{1}{2}(a + b) + \frac{1}{2}\vert a - b \vert .
\end{equation}
Therefore, for any $x \in \mathbb{R} $,
\[
\max \{ f(x), g(x) \} = \frac{1}{2}(f(x) + g(x)) + \frac{1}{2}\vert f(x) - g(x) \vert,
\]
and hence the claim is true.
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To show that $\min \{ f,g \} = -\max \{ -f, -g \} $, we use the result from part (b). Note that
\begin{align*}
\max \{ -f, -g \} & = \frac{1}{2}(-f + (-g)) + \frac{1}{2}\vert -f - (-g) \vert \\
& = \frac{1}{2}(-f - g) + \frac{1}{2}\vert g - f \vert \\
& = \frac{1}{2}(-f - g) + \frac{1}{2}\vert f - g \vert.
\end{align*}
Therefore,
\begin{align*}
-\max \{ -f, -g \} & = -\left( \frac{1}{2}(-f - g) + \frac{1}{2}\vert f - g \vert \right) \\
& = \frac{1}{2}(f + g) - \frac{1}{2}\vert f - g \vert \\
& = \min \{ f, g \}.
\end{align*}
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Using the results from parts (a) and (b), we can show that if $f$ and $g$ are continuous at $x_0 \in \mathbb{R} $, then $\min \{ f,g \} $ and $\max \{ f,g \} $ are continuous at $x_0$. Since $f$ and $g$ are continuous at $x_0$, the functions $f(x) + g(x)$ and $f(x) - g(x)$ are also continuous at $x_0$. This implies $\vert f(x) - g(x) \vert $ is also continuous. Therefore, $\min \{ f(x), g(x) \}$ and $\max \{ f(x), g(x) \}$, being sums of continuous functions, are continuous at $x_0$.