Which point of the graph of $y=\sqrt{x}$ is closest to the point $(1,0)$? This problem was assigned for an AP Calculus AB class and was not allowed a calculator:
Which point of the graph of $y=\sqrt{x}$ is closest to the point $(1,0)$?
We are not given answers and the teacher will be absent for $2$ weeks. I need to check my answer $(1/2,1/4)$ before she returns.
Here is my work:
$d=\sqrt{(x-1)^2+(y-0)^2}$
$d=\sqrt{(x-1)^2+x}$
$d^2=(x-1)^2+x$
$2dd'=2(x-1)+1$
$d'=(2(x-1)+1)/(2d)$
$0=2(x-1)+1$
$x=1/2$

As mentioned in the comments, the answer is
$$\left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)$$
(or you can use $\frac{\sqrt{2}}{2}$ for the $y$-coordinate, they're the same.) Remember that you are looking for a point on the curve, and so need to substitute the $x$-value into the function describing that curve in order to get the corresponding $y$-coordinate, and here, that means taking the square root.
The corresponding minimizing distance is, of course, $\frac{\sqrt{3}}{2}$ units, or about 0.8660 (like $\sin 60^\circ$).
