What is the point on the graph of y= (x+1)^.5 closest to the point (3,0)?

Question

Optimization Problem

Answer 1

Let (x,y) be the point that is closest to (3,0).
The distance between (x,y) and (3,0) is
d = sqrt((x-3)^2 + (y-0)^2)
Optimizing d is the same as optimizing d^2, so you want to optimize
d^2 = (x-3)^2 + y^2
Note that y = (x+1)^(1/2), so replace y with (x+1)^(1/2)
d^2 = (x-3)^2 + (x+1)
Take the derivative with respect to x.
(d/dx)d^2 = 2(x-3) + 1
Set it equal to zero.
2(x-3) + 1 = 0
2x - 6 + 1 = 0
2x - 5 = 0
2x = 5
x = 5/2 (don't forget to verify that this is a min.)
y = sqrt(5/2 + 1)
y = sqrt(7/2)
y = sqrt(14)/2
Answer: (x,y) = (5/2, sqrt(14)/2)
If you want the distance.
d^2 = (5/2 - 3)^2 + (5/2 + 1)
d^2 = (-1/2)^2 + (7/2)^2
d^2 = 1/4 + 49/4
d^2 = 50/4
d = sqrt(50/4)
d = 5sqrt(2)/2

Answer 2

The distance from (3,0)
minimize(x-3)^2+y^2 and take square root
y^2 = x+1 given
so it is (x-3)^2+x+1
= x^2-5x+10
differentiate and equate to 0
2x = 5
or x = 5/2

y^2 = (5/2+1) = 7/2
y = sqrt(7/2)
so the point is (5/2, sqrt(7/2))

Answer 3

not sure