Let D be the distance from a point on the function line from (16, 1/2)
D^2 = (x-16)^2 + (y - 1/2)^2
= (x-16)^2 + (x^2 -1/2)^2
When this is a minimum, dD^2/dx = 0
2(x-16)+ 2(x^2 -1/2)*2x = 0
2x - 32 + 4x^3 -2x = 0
4x^3 = 32
x = +/- 2
y = 4
Find the point on the graph of the function that is closest to the given point.
Function Point
f(x) = x^2 (16, 1/2)
(x, y) = ( ) ?
2 answers
Thanks :)