Question

Find the point(x,y) on the graph of y=√x nearest the point(4,0)

Answer 1

F(x)=sqrt(x)
D(x)=sqrt((f(x)-0)^2+(x-4)^2)  [distance of (4,0) from any point on graph]
=sqrt((√x-0)^2+(x-4)^2)
=sqrt(x+(x-4)^2)
To minimize distance, we equate D'(x)=0, or equivalently, we equate the derivative of D(x)^2 to zero.
The latter is easier to derive, and gives the same results.
We will do both.
D'(x)=(1/2)(2(x-4)+1)/sqrt(x+(x-4)^2)   [using chain rule]
D'(x)=0 => (2(x-4)+1)=0 => x-4=-1/2 => x=7/2

d(D²(x))/dx
=d(x+(x-4)^2)/dx
=2(x-4)+1
=> 2(x-4)+1=0  => x=7/2
So the point on the graph is (sqrt(7/2),7/2)=(1.8708,3.5)  approx.