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 October 20th, 2009, 01:51 PM #1 Member   Joined: Sep 2009 Posts: 51 Thanks: 0 Newton's Method Find the point on the graph of f(x) = (x + 3)^2 + 6 that is closest to the point (1, 3). Use Newton's Method to approximate the solution to 2 decimal places. (__, __) I couldn't figure it out that way so I tried linear approximation, and still i don't think i did it the correct way. f(c)+f'(c)(x-c) f ' (x) =2x+6 y-f(1)= f '(1)(x-1) y-22=8(x-1) y=8x+14 so I got... (1,22)
 October 20th, 2009, 03:21 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond Re: Newton's Method Set up the distance equation $D\,=\,\sqrt{(x\,-\,1)^{2}\,+\,(\,(\,(x\,+\,3)^{2}\,+\,6)\,-\,3)^{2}}$. Calculate the derivative (hint: you only need to take the derivative of the expression under the root sign). To use Newton's Method keep in mind that you are given the equation of a parabola in vertex form, and the point (1, 3) is to the right of this.
 October 20th, 2009, 03:56 PM #3 Senior Member   Joined: Mar 2007 Posts: 428 Thanks: 0 Re: Newton's Method Although you should use Newton's method if asked, you can double check by finding the expression for the square of the distance from curve to point [already done for you] and differentiate as suggested then set to zero for max/min. The problem then is a cubic equation with only one real root, which can be found graphically. ..But again, this is only to double check what to expect and compare with your result. Of course, the value found would have to be put back into the expression for distance to find that.

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