First, thank you for reading my post!
I have been trying to solve the following problem(image attached below) and I have zero idea on how to approach this. I am self reading the book and was able to figure out the first 7 problems of this problem set. Can someone please give me an idea how to start this problem?
My approach: We would want $|x_n-\alpha|<\epsilon,$ for arbitrary $\epsilon<0$ and large enough $n.$ This is equivalent to making $|g(x_n)-cf(x_n)-\alpha|<\epsilon.$ From here, I am clueless as how to proceed or how to use the given data.
Thank you in advance!
Root is $\alpha$. $x_n$ near $\alpha$, so $0<|x_n-\alpha|<\delta$
$\frac{|x_n+cf(x_n)-\alpha|}{|x_n-\alpha|}<1$. Condition for convergence.
$\frac{|x_n-\alpha+cf(x_n)-cf(\alpha)|}{|x_n-\alpha|}<1$ Adding zero to numerator.
$|x_n-\alpha+cf(x_n)-cf(\alpha)|<|x_n-\alpha|$
$|x_n-\alpha||1+c\frac{f(x_n)-f(\alpha)}{x_n-\alpha}|<|x_n-\alpha|$. Factor out $(x_n-\alpha)$
$|1+c \frac{f(x_n)-f(\alpha)}{x_n-\alpha}|<1$. Divided.
$-1<1+c\frac{f(x_n)-f(\alpha)}{x_n-\alpha}<1$. Re-expessed as a complex inequality.
$-2<c\frac{f(x_n)-f(\alpha)}{x_n-\alpha}<0$
$-2< cf'(\alpha)<0$ Condition on $c$ with derivative.
We don't know if $f'(\alpha)$ is positive or negative so:
$|c|<\frac{2}{|f'(\alpha)|}$.