> 
> Show $x\_{n+1}=x\_n-\frac{f(x\_n)}{f'(x\_n)}$ with start value $x\_0=1$ and $f:x\to x^5$ does not converge quadratically. 
> 
> 
> 

So by definition I have to show that $\mid x\_{n+1}-0\mid\le c \mid x\_n -0\mid^2$ for quadratic convergence, I guess.
Plugging in leads to $\mid x -\frac{f(x)}{f'(x)}\mid=\mid x-\frac x5\mid=\frac45 \mid x\mid\le c \mid x \mid^2$ And since $x$ will get never negative in this iteration we can write above as $\frac45 x\le c x^2 \Rightarrow \frac45 \le cx$, but with this reasing I could set $c=\frac 1x$ and therefore this convergence would be quadratic. I am oviously on the wrong track. Could someone help me and guide me, what exactly I have to do here?