<s>[INST] Calculating the shortest possible distance between points Question:
Given the points $A(3,3)$, $B(0,1)$ and $C(x,0)$ where $0 < x < 3$, $AC$ is the distance between $A$ and $C$ and $BC$ is the distance between $B$ and $C$. What is x for the distance $AC + BC$ to be minimal?
What have I done?
I defined the function $AC + BC$ as:
$\mathrm{f}\left( x\right) =\sqrt{{1}^{2}+{x}^{2}}+\sqrt{{3}^{2}+{\left( 3-x\right) }^{2}}$
And the first derivative:
$\mathrm{f'}\left( x\right) =\frac{x}{\sqrt{{x}^{2}+1}}+\frac{3-x}{\sqrt{{x}^{2}-6\,x+18}}$
We need to find the values for $\mathrm{f'}\left( x\right) = 0$, so by summing and multiplying both sides I got to the equation:
$2x^4 - 12x^3  + 19x^2 -6x+18 = 0$
But I don't think the purpose should be to solve a 4th grade equation, there should be another way I'm missing..
 [/INST] Norbert and Ross Millikan have already suggested one slick solution, and Gerry Myerson the even slicker solution, but you can do it purely algebraically: you ran into trouble because your derivative isn’t quite right. It should be
$$f\,'(x)=\frac{x}{\sqrt{1+x^2}}-\frac{3-x}{\sqrt{18-6x+x^2}}\;,$$ 
with the second term negative from the chain rule applied to $3-x$. Setting that to $0$ and doing a little algebra, we get
$$\begin{align*}
x\sqrt{18-6x+x^2}&=(3-x)\sqrt{1+x^2}\;,\\
18x^2-6x^3+x^4&=(3-x)^2(1+x^2)\;,\\
18x^2-6x^3+x^4&=9-6x+10x^2-6x^3+x^4\;,\\
18x^2&=9-6x+10x^2\;,\\
8x^2+6x-9&=0\;,\text{ and}\\
(4x-3)(2x+3)&=0\;.
\end{align*}$$
The desired solution is clearly $x=\dfrac34$.
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