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..

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 $x=\dfrac34$.
