1. ## Newton's Method Help...

I need help with this problem:

Any ideas?

2. Hi,
the first thing to do is to convert the above question to a root finding problem.

you can do so by first finding distance of an arbituary point on the curve from (-3,0).

you should get the equation of d = sqroot( (Xo - (-3) )^2 + (f(Xo) - 0 )^2 ) , where (Xo, f(Xo)) is an arbituary point on the curve.

since distance in this case has an absolute minimum, use Newton's Method to find the root of this derivative, d'(Xo) = 0, to solve for the value of Xo where this absolute minimum occurs, ie, the point closest to (-3, 0).

Hope this helps!

3. Originally Posted by BeSweeet
I need help with this problem:

Any ideas?
Also, we can solve for the square of the distance being a minimum.

$\displaystyle (x+3)^2+y^2=minimum$

$\displaystyle (x+3)^2+\left[(x+7)^2+3\right]^2=minimum$

Differentiating and equating to zero

$\displaystyle 2(x+3)+2\left[(x+7)^2+3\right]2(x+7)=0$

$\displaystyle 4x^3+84x^2+602x+1462=0$
$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$
$\displaystyle f'(x)=12x^2+168x+602$