What we are being asked here is to simply minimize distance. Also, note that we can write #f(x)=sqrt(x)# as #y=sqrt(x)#.
Now, what is this "distance?" How do we find it? Well, if you think back to Algebra I or Geometry, you'll remember that the distance between two points #(x_1,y_1)# and #(x_2,y_2)# is given by: #sqrt((y_2-y_1)^2+(x_2-x_1)^2)#. For example, the distance between the points #(4,0)# and #(0,3)# would be:
#sqrt((3-0)^2+(4-0)^2)=sqrt(9+16)=sqrt(25)=5#
Ok, so what is #(x_1,y_1)# and #(x_2,y_2)# in our example? #(x_1,y_1)# is simple - it's just the point given in the problem, #(4,0)#. Because we don't know what #x_2# is, we'll just call it #x# for now. As for #y_2#, we don't know that either; and since #y=sqrt(x)#, we'll call it #sqrt(x)#.
Our formula then becomes:
#sqrt((sqrt(x)-0)^2+(x-4)^2)=sqrt((sqrt(x)^2)+x^2-8x+16)=sqrt(x+x^2-8x+16)=sqrt(x^2-7x+16)#
We are being asked to minimize this distance, which we'll call #s# to make the following calculations easier. To minimize something, we have to take its derivative, so let's start there:
#s=sqrt(x^2-7x+16)=(x^2-7x+16)^(1/2)#
#(ds)/dx=(2x-7)*1/(2(x^2-7x+16)^(1/2))->#Using power rule and chain rule
#(ds)/dx=(2x-7)/(2sqrt(x^2-7x+16)#
Now we set this equal to #0# and solve for #x#:
#0=(2x-7)/(2sqrt(x^2-7x+16)#
#0=2x-7#
#x=7/2#
This is known as the critical value, and it represents the #x#-value for which the function is minimized. All we need to do now is find the corresponding #y#-value, using the definition of #y#: #y=sqrt(x)#. Substituing #7/2# for #x#:
#y=sqrt(7/2)#
#y~~1.87#
And voila, the #y#-value. We can now say that the minimum distance between #f(x)=sqrt(x)# and the point #(4,0)# (the place where these two are closest) occurs at #(7/2, 1.87)#. For a little extra fun, we can use the distance formula to see what the actual distance between the points is:
#s=sqrt((1.87-0)^2+(7/2-4)^2)~~1.8# units
