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 $x=1/2$, $y=\sqrt{1/2}$.
