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March 6, 2015

March 6, 2015

Posted by **Malika** on Sunday, March 4, 2012 at 9:51pm.

Function Point

f(x) = x^2 (16, 1/2)

(x, y) = ( ) ?

- Math -
**drwls**, Monday, March 5, 2012 at 8:52amLet 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

- Math -
**Malika**, Monday, March 5, 2012 at 11:27amThanks :)

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