find the point on the curve y=sqr.(x) that is the closest to the point (3,0).

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This is a minimization problem hence, you need to find a function that models the distance between a point `A(x,y)` that lies on the given curve and the fixed point `(3,0).`

You need to use the distance formula such that:

`d = sqrt((x - 3)^2 + (y - 0)^2)`

You need to write the distance formula in terms of one variable, hence, you need to use the equation of the curve to write y in terms of x such that:

`y = sqrt x => y^2 = x`

Substituting x for `y^2` in equation of distance yields:

`d = sqrt((x - 3)^2 + x) => d = sqrt(x^2 - 6x + 9 + x)`

`d(x) = sqrt(x^2 - 5x + 9)`

You need to find derivative of this function such that:

`d'(x) = ((x^2 - 5x + 9)')/(2sqrt(x^2 - 5x + 9))`

`d'(x) = (2x - 5)/(2sqrt(x^2 - 5x + 9))`

You need to solve the equation `d'(x) = 0` to find the extreme values of the distance function such that:

`(2x - 5)/(2sqrt(x^2 - 5x + 9)) = 0 => 2x - 5 = 0 => 2x = 5= > x = 5/2`

You need to find y coordinate at `x = 5/2` such that:

`y = sqrt x => y = sqrt(5/2) => y = (sqrt10)/2`

**Hence, evaluating the point that lies on the curve `y = sqrt x` and it is closest to the fixed point (3,0) yields`A(5/2 ; (sqrt10)/2).` **

**Sources:**

Any point on the curve is (x,y) = (x, sqrtx).

The distance d between the points (3,0) and (x, sqrtx) is given by:

d^2 = (3-x)^2+(sqrtx-0)^2.

d^2 = 9 -2sqrtx+x+x

d^2 = 9 - 2sqrtx +2x..(1)

When d is minimum, d^2 = is also minimum.

d^2 to be minimum, (9 - 2sqrtx +2x)' = 0 and (9 - 2sqrtx +2x)'' > 0.

=> -2/(2sqrtx) +2 = 0

=> -x^(-1/2) +1 = 0

=> x = 1.

(9 - 2sqrtx +2x)'' = (-)(-1/2)x^(-3/2) = (1/2)x^(-3/2) = 1 at x= 1.

Put x = 1 in (1), d^2 = (9-2sqrt1+2*1)

d^2 = 9-2+2 = 9

d = 9^(1/2) = 3.

**So 3 is the shortest distance between the curve y = sqrtx and the point (3,0)**.

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