Determine the point on the curve y = 1 + x^(3/2) that are closest to the point (x,y) = (4,1)

Let d be the distance between a point (x, 1 + x^(3/2) ) on the curve and the point of (4,1)
d^2 = (x-4)^2 + (1 + x^(3/2) - 1)^2 = (x-4)^2 + x^3
Take derivative with respect to x,
2dd' = 2(x-4) + 3x^2
Solve d' = 0 for x,
3x^2 + 2x - 8 = (3x - 4)(x + 2) = 0
Answer: x = 4/3, y = 1 + (4/x)^(3/2)

This one is easy. You have to do a point product between a generalized vector that starts in the point given and the curve. As you know that in Euclidean space is a straight line and that the shortest distance between a point and a curve is perpendicular (ortogonal) to the curve, the point product must be equal to 0. After that you can determine the components of the ending part of the vector and that will give you (along with the starting components (4,1)) the vector that reaches to the curve in its closest point to (4,1).