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.

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:
$$(d(x)^2)'=\left((x-1)^2+x\right)'=2(x-1)+1\stackrel{?}=0\iff 2(x-1)=-1\ldots etc.$$
