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added exact solution
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@ -307,8 +307,42 @@ As you can easily verify, only $x_1$ is a minimum of $d_{P,f}$.
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It is obvious that a quadratic function can have two points with
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minimal distance.
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For example, let $f(x) = x^2$ and $P = (0,5)$. Then $P_{f,1} \approx (2.179, 2.179^2)$
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has minimal distance to $P$, but also $P_{f,2}\approx (-2.179, 2.179^2)$.\todo{exact example?}
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For example, let $f(x) = x^2$ and $P = (0,5)$. Then $P_{f,1} = (\sqrt{\frac{9}{2}}, \frac{9}{2})$
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has minimal distance to $P$, but also $P_{f,2} = (-\sqrt{\frac{9}{2}}, \frac{9}{2})$.
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\begin{figure}[htp]
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\centering
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\begin{tikzpicture}
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\begin{axis}[
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%legend pos=north west,
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axis x line=middle,
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axis y line=middle,
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grid = major,
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width=0.6\linewidth,
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height=8cm,
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grid style={dashed, gray!30},
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xmin=-3, % start the diagram at this x-coordinate
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xmax= 3, % end the diagram at this x-coordinate
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ymin= 0, % start the diagram at this y-coordinate
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ymax= 5, % end the diagram at this y-coordinate
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axis background/.style={fill=white},
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xlabel=$x$,
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ylabel=$y$,
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%xticklabels={-2,-1.6,...,7},
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%yticklabels={-8,-7,...,8},
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tick align=outside,
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minor tick num=-3,
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enlargelimits=true,
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tension=0.08]
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\addplot[domain=-3:3, thick,samples=50, orange] {x*x};
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\draw (axis cs:0,5) circle[radius=2.17];
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\draw[red, thick] (axis cs:0,5) -- (axis cs:2.121,4.5);
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\draw[red, thick] (axis cs:0,5) -- (axis cs:-2.121,4.5);
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\addlegendentry{$f(x)=x^2$}
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\end{axis}
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\end{tikzpicture}
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\caption{Two points with minimal distance}
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\end{figure}
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As discussed before, there cannot be more than 3 points on the graph
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of $f$ next to $P$.
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