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\chapter{Linear function}
\section{Defined on $\mdr$}
Let $f(x) = m \cdot x + t$ with $m \in \mdr \setminus \Set{0}$ and
$t \in \mdr$ be a linear function.
\begin{figure}[htp]
\centering
\begin{tikzpicture}
\begin{axis}[
legend pos=north east,
axis x line=middle,
axis y line=middle,
grid = major,
width=0.8\linewidth,
height=8cm,
grid style={dashed, gray!30},
xmin= 0, % start the diagram at this x-coordinate
xmax= 5, % end the diagram at this x-coordinate
ymin= 0, % start the diagram at this y-coordinate
ymax= 3, % end the diagram at this y-coordinate
axis background/.style={fill=white},
xlabel=$x$,
ylabel=$y$,
tick align=outside,
minor tick num=-3,
enlargelimits=true,
tension=0.08]
\addplot[domain=-5:5, thick,samples=50, red] {0.5*x};
\addplot[domain=-5:5, thick,samples=50, blue] {-2*x+6};
\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
\addlegendentry{$f(x)=\frac{1}{2}x$}
\addlegendentry{$g(x)=-2x+6$}
\end{axis}
\end{tikzpicture}
\caption{The shortest distance of $P$ to $f$ can be calculated by using the perpendicular}
\label{fig:linear-min-distance}
\end{figure}
Now you can drop a perpendicular $f_\bot$ through $P$ on $f(x)$. The
slope of $f_\bot$ is $- \frac{1}{m}$ and $t_\bot$ can be calculated:\nobreak
\begin{align}
f_\bot(x) &= - \frac{1}{m} \cdot x + t_\bot\\
\Rightarrow y_P &= - \frac{1}{m} \cdot x_P + t_\bot\\
\Leftrightarrow t_\bot &= y_P + \frac{1}{m} \cdot x_P
\end{align}
The point $(x, f(x))$ where the perpendicular $f_\bot$ crosses $f$
is calculated this way:
\begin{align}
f(x) &= f_\bot(x)\\
\Leftrightarrow m \cdot x + t &= - \frac{1}{m} \cdot x + \left(y_P + \frac{1}{m} \cdot x_P \right)\\
\Leftrightarrow \left (m + \frac{1}{m} \right ) \cdot x &= y_P + \frac{1}{m} \cdot x_P - t\\
\Leftrightarrow x &= \frac{m}{m^2+1} \left ( y_P + \frac{1}{m} \cdot x_P - t \right )
\end{align}
There is only one point with minimal distance. See Figure~\ref{fig:linear-min-distance}.
\section{Defined on a closed interval of $\mdr$}