Find the minimum distance from the point Find the minimum distance from the point $ \ (5,0) \ $ to the curve $ \ y=\sqrt x+2 \ $. 
Answer:
Let $ \ (x,y) \ $ be the closest point on $ \ y=\sqrt x+2 \ $ from $  (5,0) $ . 
Then the distance is given by 
$ d(x,y)=\sqrt{(x-5)^2+y^2} \ $ 
We will minimize the function $ \ g(x,y)=(x-5)^2+y^2 \ $ 
replacing $ y \ \ by \ \ \sqrt x+2 \ $ , we get 
$ g(x)=(x-5)^2+(\sqrt x+2)^2 \ $ 
The extreme points \ are 
$ f'(x)=0 \\  2(x-5)+\frac{\sqrt x+2}{\sqrt x} =0 \\ 2x \sqrt x-9 \sqrt x+2=0 $ 
This becomes complicated . I am  unable to calculate the closest point.
Help me out

Now, let $\sqrt{x}=t$.
Thus,$$2x\sqrt{x}-9\sqrt{x}+2=2t^3-9t+2=2t^3-4t^2+4t^2-8t-t+2=$$
$$=(t-2)(2t^2+4t-1).$$
Thus, $x_{min}=4$ and $x_{max}=\left(\sqrt{1.5}-1\right)^2.$
Also, we need to check, what happens for $x=0$.
