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:
$d=\sqrt{(x-1)^2+(y-0)^2}$
$d=\sqrt{(x-1)^2+x}$
$d^2=(x-1)^2+x$
$2dd'=2(x-1)+1$
$d'=(2(x-1)+1)/(2d)$
$0=2(x-1)+1$
$x=1/2$
 A: 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 $x=1/2$, $y=\sqrt{1/2}$.
A: An algebra-free approach:

In order to draw the tangent from a point $P$ on a parabola, it is sufficient to project $P$ on the axis, reflect this point with respect to the vertex and join the new point with $P$. Since the tangent drawn from $P=\left(\frac{1}{2},\frac{1}{\sqrt{2}}\right)$ is orthogonal to the line joining $P$ with $(1,0)$ (by Euclid's second theorem on right triangles, or just by computing slopes), $P$ is the wanted solution.
A: 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 $\sin 60^\circ$).
A: We may look at differently: 
Let the point we are after be $P:(x_0,\sqrt x_0)$. Now say we find the equation of the line, $l_1$, that passes through $P$ and $(1,0)$. Geometrically, $l_1$ is perpendicular to a tangent line to $y=\sqrt x$ that passes through $P$, hence its slope is $-\left(\frac1{2\sqrt x_0}\right)^{-1}$ and as such$$l_1:y=-2\sqrt x_0 (x-1).$$
Now noting that $l_1$ also passes though $P$ we have $$\sqrt x_0=-2\sqrt x_0 (x_0-1),$$which implies $x_0=\frac12$.
A: The solution point is such that the circle centered at $(1,0)$ tangents the curve, i.e. the system of equations
$$\begin{cases}(x-1)^2+y^2=r^2,\\y=\sqrt x\end{cases}$$ has a double root. By eliminating $y$,
$$(x-1)^2+x-r^2=0$$ has a double root when $$2(x-1)+1=0,$$ hence $$\left(\frac12,\frac1{\sqrt2}\right).$$
A: 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 $d(x_0)= \min\{d(x): x \ge 0 \}.$
