First note that the distance between an arbitrary point #(x,y)=(x,sqrt(x))# on the graph of #y=sqrt(x)# and the point #(4,0)# is #sqrt((x-4)^2+(sqrt(x)-0)^2)=sqrt(x^{2}-7x+16)#.

Next, note that the value of #x# where #d(x)=sqrt(x^{2}-7x+16)# is minimized is the same value of #x# where #s(x)=(d(x))^2=x^2-7x+16# is minimized.

Since #s'(x)=2x-7#, #s(x)# and #d(x)# are minimized at #x=7/2=3.5# (note that #s''(x)=2>0# so that the critical point is a minimum).

Therefore, the point on the graph of #y=sqrt(x)# that is closest to #(4,0)# is #(7/2,sqrt(7/2))\approx (3.5,1.87083)#. The minimum distance is #d(7/2)=sqrt(49/4-49/2+16)=sqrt(15/4)=sqrt{15}/2\approx 1.93649#