Finding the point on $f = \sqrt x$ closest to $(1,0)$ Find the point on $y = \sqrt x$ closest to $(1,0)$. I tried setting it $-x=x^{1/2}$ and then solve but I really am confused. please help.
 A: In calculus, you have a strong method optimizing functions, i.e. finding minima and maxima (Do you know what method I'm talking about?). here, you have a curve $y = \sqrt x$. But this is not the function you are optimizing. You are looking for the closest point on the curve to $(1,0)$.
This means you want to minimize the function that gives the distance from a point on your curve to the point $(1,0)$.
So your goals should be something like this:


*

*Write down explicitly the function for the distance from a point on your curve to $(1,0)$

*Optimize it, likely involving a derivative

A: Hints:
Take a point $\,(x,\sqrt x)\, $ on $\,y=\sqrt x\,$ , so its distance from $\,(1,0)\,$ is
$$d(x):=\sqrt{(x-1)^2+x}$$
Now you have to find the maximal value of the above. You can take $\,d(x)^2\,$ since it's easier and yields the same maximal point (why?) , so differentiating:
$$(d(x)^2)'=\left((x-1)^2+x\right)'=2(x-1)+1\stackrel{?}=0\iff 2(x-1)=-1\ldots etc.$$
A: Hint:  Let $(x,y)$ be a point on $y=x^{1/2}$. Then minimize the square of the distance between $(1,0)$ and $(x,y)$.
A: How would you calculate the distance between the point $(x,\sqrt x)$ and $(1,0)$?  For what value of $x$ is this distance a minumum?
