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 a check:

Partial answer, completing the square.

$x,y\ge 0$.

$d^2= (x-1)^2+x= $

$x^2-2x+1+x= x^2-x+1;$

$d^2=(x-1/2)^2 -1/4+1 =$

$(x-1/2)^2 +3/4\ge 3/4$ (why?).

$d^2_{min} =3/4$, at $x=1/2$, $y=\sqrt{1/2}$.
Let $(x, \sqrt{x})$ a point on the graph. The distance to$(1,0)$ is then given by

$d(x)=((x-1)^2+\sqrt{x}^2)^{1/2}=((x-1)^2+x)^{1/2}$.

Let $f(x):=d(x)^2.$

It is now your turn to determine $x_0 \ge 0$ such that $f(x_0)= \min\{f(x): x \ge 0 \}.$

(we get $x_0 =1/2$).

We then have $d(x_0)= \min\{d(x): x \ge 0 \}.$