First note that the distance between an arbitrary point (x,y)=(x,sqrt(x))(x,y)=(x,√x) on the graph of y=sqrt(x)y=√x and the point (4,0)(4,0) is sqrt((x-4)^2+(sqrt(x)-0)^2)=sqrt(x^{2}-7x+16)√(x−4)2+(√x−0)2=√x2−7x+16.
Next, note that the value of xx where d(x)=sqrt(x^{2}-7x+16)d(x)=√x2−7x+16 is minimized is the same value of xx where s(x)=(d(x))^2=x^2-7x+16s(x)=(d(x))2=x2−7x+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