How do you find the point on the the graph #y=sqrtx# which is plosest to the point (4,0)?

Question

Isabella Ackerman · Accepted Answer

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#

Axel Alvarez · Answer

undefined To find the point on the graph y = √x closest to the point (4, 0), you need to minimize the distance between the two points.

1. Begin by expressing the distance between two points (x, y) and (4, 0) in terms of the variables x and y.
2. Square the distance formula to eliminate the square root.
3. Differentiate the squared distance function with respect to x.
4. Set the derivative equal to zero and solve for x.
5. Substitute the value of x obtained into the equation y = √x to find the corresponding y-coordinate.
6. The point (x, y) obtained is the point on the graph y = √x closest to the point (4, 0).