Convergence of sequence involving fixed points of a function

Question

It is given that the function $f(x) = \frac{x^3 + 1}{3}$ has three fixed points $\alpha$,$\beta$,$\gamma$ in (-2,-1), (0, 1) and (1, 2) respectively. A sequence of real numbers is defined as $$x_1 =\gamma - 0.01\qquad , \qquad x_{n+1} = f(x_n) \enspace\forall n = 1, 2, 3,\cdots.$$ Given that the sequence converges, find $\displaystyle\lim_{n\rightarrow\infty}x_n$.

Answer 1

Since the derivative of your function $$y'(x) = x^2$$ is greater than $1$ at the fixed point $x=\gamma $, the fixed point is an unstable one and since your $y''(x)$ is positive a staring point of $\gamma -0.1$ will move to the left and get attracted to the stable fixed point $\beta$ in $(0,1)$