For $x^3−2x^2−13x+30 = 0$, with root r = 3. I am supposed to add $cx$ to both sides of the equation before dividing by $c$ to obtain the fixed point equation $$g(x)=x$$ where $$g(x) = \frac{1}{c}x^3−\frac{2}{c}x^2−\frac{13-c}{c}x+\frac{30}{c}$$ I was then told to find the values of c such that the fixed point iteration locally convergent to r = 3.
My idea is to differentiate g(x) to get $g'(x) = \frac{3}{c}x^2-\frac{4}{c}x+\frac{c-13}{c}$. For fixed point iteration to locally converge, $|g'(x)| < 1$. However, when trying to solve $|g'(x)| < 1$, I realise there are both $x$ and $c$ present. Am I supposed to use $b^2 - 4ac > 0$ to solve this problem?