y=x^(1/2). what's the shortest distance from that curve to the point (1/2,0)?
Let $\displaystyle \left(x,x^{\frac{1}{2}} \right)$ be an arbitrary point on the curve. It will be simpler to minimize the square of the distance between this arbitrary point and the given fixed point. Can you state the square of this distance?
We have two points:
$\displaystyle \left(x,x^{\frac{1}{2}} \right)$ and $\displaystyle \left(\frac{1}{2},0 \right)$.
Now, the distance formula tells us the distance $\displaystyle d$ between the two points $\displaystyle P_1(x_1,y_1)$ and $\displaystyle P_2(x_2,y_2)$ is:
$\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
So, choose either of our two points to be $\displaystyle P_1$ and the other as $\displaystyle P_2$ and plug into the equivalent:
$\displaystyle d^2=(x_2-x_1)^2+(y_2-y_1)^2$
Then simplify and optimize by differentiation. Be sure to demonstrate you have minimized by using an appropriate test.