Is $\frac{1}{x}$ continuosly differentiable on the set $E = (- \infty, 0) \cup (0, \infty)$?

Question

Just to explain the motivation behind some other thing, I shall ask first this question:

Is $\frac{1}{x}$ continuosly differentiable (or we can examine $\frac{1}{x^3}$ also) on the set $$E = (- \infty, 0) \cup (0, > \infty)$$

The reason why I'm asking this is that when we consider a solution to a given initial value problem, we define the definition of interval of that solution as a single connected set (or just an interval), and the reason for this is explained to me as that we are trying to find solutions $y(x) \in C^1$, but when I discard the problematic points from my domain, why should I get any problem about the condition that $y \in C^1$ ?

Edit:

For example, let say that we solve an ODE, and the general solution is of the form $$y(x) = c / x.$$ Then if the initial condition is given as $y(1) = 1$, we see that $c = 1$, and the domain of definition is $1 \in (0,\infty)$. However, I do not understand the motivation behind the reason why the interval of definition is not $(-\infty, 0) \cup (0, \infty)$ .

Answer 1

Yes, $1/x$ and $1/x^3$ are continuously differentiable (in fact, infinitely differentiable) on your domain (the set of all nonzero reals).

The reason we typically assume that the domain of a solution to an IVP is connected is to ensure uniqueness. If you try to solve the IVP $f'(x)=-1/x^2$ with $f(1)=1$ on your domain, the solution isn't unique: the values of the function aren't determined in the left half-plane.

Answer 2

The funciton $f(x)=\frac1x$ is an element of $C^1(\mathbb R\setminus\{0\})$.

To comment on your question about the initial value problems, I would need more information about what is unclear to you.

Answer 3

Your notation is inadequate. The function is in $C^1(E)$. It is also in $C^1(U)$ for every interval $U\subset E$. You must specify the domain in this notation if it isn’t implicit, especially when you are considering several of them. In other words, $C^1$ without a specified domain is ambiguous.