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I want to show that if a function is linear for example F(x) = 3x + 3, then after one iteration of Newton's method I can find the x-value such that F(x) = 0. In this case, after one iteration x = -1. I see that it works after one iteration, but how do I prove that??

David Zheng
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    Be careful in your definition of linear. Usually a linear function is defined to be one for which $f(x+y)=f(x)+f(y)$, which is not true for your function. Do you mean functions whose graphs are lines (IE linear polynomials)? – Alan Feb 11 '21 at 16:55
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    @Moo thanks that is exactly what I am looking for – David Zheng Feb 11 '21 at 17:06

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Usual setup for Newton's method:

$$f'\left(x_0\right) = \lim_{x \rightarrow x_0} \frac{f\left(x\right) - f\left(x_0\right)}{x-x_0}$$

Look for intersections, i.e. $f\left(x\right)=0$:

$$x = x_0 - \frac{f\left(x_0\right)}{f'\left(x_0\right)}$$

Now stray from the general case and suppose $f\left(x\right)$ is linear, as in $f\left(x\right) = mx+b$, so $f'\left(x\right) = m$. Then,

$$x = x_0 - \frac{mx_0 + b}{m} = x_0 - x_0 - \frac{b}{m} = - \frac{b}{m}$$

... and you have just $-b/m$ after one iteration.