Maximum and minimum help. :D?

Question

Which point on the curve y=sqrt of x-2 +1 is closest to the point (4,1)? What is the minimum distance?

*x-2 is the only one with the squareroot.

thanks to those who will answer. =]

Answer 1

Well there are an infinite number of points on the graph y = sqrt(x-2) + 1.  So which point is the minimum?  Well lets designate the minimum point as (x,y) where y = sqrt(x-2) + 1

So we have the point:

(x, sqrt(x-2) + 1)

This point wants to be the closest to (4,1)

So we will use the distance formula:

D = sqrt((x2-x1)^2 + (y2-y1)^2)

D = sqrt((4-x)^2 + (sqrt(x-2) + 1 - 1)^2)

D = sqrt(16 - 8x + x^2 + x - 2)

D = sqrt(14 - 7x + x^2)

Well we want this distance to be a minimum, so we must differentiate this function:

D' = (1/2)(-7 + 2x)/sqrt(14-7x+x^2)

Set this equal to zero:  Since we are looking for a point on the graph we will only set the numerator equal to zero:

(1/2)(2x-7) = 0

2x-7 = 0

2x = 7

x = 7/2

Now take this value, plug it back into you y = equation to find the y-value and then take both x and y and use the distance formula to find D.  i'll leave that part for you.

Voila

Note:  your teacher may ask you to prove that x = 7/2 is a minimum.  If so, use the first derivative test to prove in fact it is a minimum :)