<s>[INST] Find the point on the graph of $f (x) = \sqrt x$ closest to the point $(3,0)$ $(x,y)=(x,\sqrt{x})$ 
$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
$=\sqrt{x^4-5x^2+9}$
$g(x)=x^4-5x^2+9$ , $g'(x)=4x^3-10x=0$ $: x=0, x=+-\sqrt{10}/2$
The question is: Should I put those x-values in g(x) or the orginal graph,$ 
 y=\sqrt{x}$. To me, it's logical to put it into g(x), but in an example it was reverse
 [/INST] You want to minimize the function
$$
f(x)=\|(x,\sqrt{x})-(3,0)\|^2=\|(x-3,\sqrt{x})\|^2=(x-3)^2+(\sqrt{x})^2=x^2-5x+9
$$
The function $f$ reaches its minimum value at 
$$
h=-\dfrac{-5}{2(1)}=\dfrac{5}{2}
$$
and the minimum value is
$$
k=f(h)=\dfrac{11}{4}
$$
Hence the point of $y=\sqrt{x}$ closest to the point $(3,0)$ is the point
$\left(\dfrac{5}{2},\sqrt{\dfrac{5}{2}}\right)$, and its distance to the point $(3,0)$ is $\sqrt{k}=\dfrac{\sqrt{11}}{2}$.
</s>