Since this problem has already been solved by minimizing the distance function between the given point and the curve, I am going to use a different approach.
The point P on the curve `y = sqrt(x)` is closest to the point (3,0) if the line connecting the two points is perpendicular to the curve (that is, perpendicular to the tangent line to the graph of `y = sqrt(x)` at P).
The two lines are perpendicular if their slopes `m_1` and `m_2` are negative reciprocals: `m_1*m_2 = -1`.
The slope of the tangent line to the graph of `y = sqrt(x)` at the point `P` is the value of the derivative y'(x) at this point:
`y'(x) = (sqrt(x))' = 1/(2sqrt(x))`
At point `P` , `y'(x_P) = -1/(2sqrt(x_p))`. So, `m_1 =- 1/(2sqrt(x_P))`.
The line connecting the points `P ` and (3, 0) has the slope that can be found using the following slope formula:
`m_2 = (0 - y_P)/(3 - x_P) = sqrt(x_P)/(3 - x_P)`.
Plugging these into `m_1*m_2 = -1`, we get the following:
`-1/(2sqrt(x_P)) * sqrt(x_P)/(3 - x_P) = -1`
which simplifies to the following:
`1/(2(3 - x_P)) = 1`
`2(3 - x_P) = 1`
`3 - x_P = 1/2`
`x_P = 3 - 1/2 = 5/2`
So, the x-coordinate of the point closest to (3,0) is 5/2, and the y-coordinate is as follows:
`y = sqrt(5/2) = sqrt(10)/2`.
The point on the given curve closest to (3,0) is `(5/2, sqrt(10)/2)`.