$$ f(x) = \left\{\begin{aligned} &0 &&: x<- 1\\ &(1/4)(x+1)(x-1)^2 &&: -1\le x \le 1\\ &0 &&: x>1 \end{aligned} \right.$$
It is also given $x\in \Omega = [-2,2]$.
Let a function be defined as above. It is asked to determine the smoothness of this function. In other words find $m$ in $C^m (\Omega)$.
I know that $x<-1$ and $x>1$ it is continuously differentiable any number of time. It is also true that $-1 \le x \le 1$ it is continuously differentiable any number of times.
How to confirm smoothness and determine the number $m$ ? I want to find the continuity of the $m$-$th$ derivative and the maximum number $m$ for which it is possible.
By my logic it is only $C^0$ continuum. Because the first derivative is not continuous at $x=-1$. Am I right?
We must calculate the left and right derivatives of $f$ to see if the derivatives exist:
$$f_+'(x) = \lim_{h\to0^+}\frac{f(x + h)-f(x)}{h}$$
$$f_-'(x) = \lim_{h\to0^-}\frac{f(x + h)-f(x)}{h}$$
Problems will occur at $x = -1$. To calculate the right derivative $f_+'(-1)$, consider the function $g : \mathbb{R}\to\mathbb{R}$ defined as $g(x) = \frac{1}{4}(x+1)(x-1)^2$. Its derivative is $$g'(x) = \frac{1}{4}(x-1)^2+\frac{1}{2}(x-1)(x+1)$$ We can conclude:
$$f_+'(-1) = \lim_{h\to0^+}\frac{f(-1 + h) - f(-1)}{h} = \lim_{h\to0^+}\frac{g(-1 + h) - g(-1)}{h} = g'(-1) = 1$$
The left derivative of $f$ at $-1$ is of course $f_-'(-1) = \lim_{h\to0^-}\frac{f(-1 + h) - f(-1)}{h} = 0$.
Since the left and right derivative are not equal, you can conclude that the derivative $f'(-1)$ does not exist. So, $f$ is not differentiable. Since you have already established that $f$ is continuous, it follows that $f \in C^0(\Omega)$.