C vs C1 Continuous: Understanding Derivatives

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The discussion clarifies the distinction between functions that are continuous and those that have continuous first derivatives, emphasizing that F ∈ C^1(E) indicates F has a continuous first derivative. An example illustrates that while f(x) = |x| is continuous everywhere, it is not differentiable at x = 0, thus not in C^1(E). A follow-up question addresses an autonomous differential equation where the solution is shown to be constant if the function f has first derivatives defined on an open set E. The user seeks confirmation on whether derivatives must be shown on the interval I and questions the nature of the domain of f. The conversation highlights the importance of understanding the relationship between continuity, differentiability, and the implications for solutions of differential equations.
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Homework Statement



Let F be a function defined on a open set E where E \in V. F is said to be continuous at each point x \in V. Which according to my textbook at hand can be written as F \in \mathcal{C}(E)

But the expression F \in \mathcal{C}^{1}(E) is that equal to saying that F have first order derivatives defined on the set E?
 
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More than that. F \in \mathcal{C}^{1}(E) says that F has a continuous first derivative. For example, f(x)= |x| is continuous in any neighborhood of x= 0 but not differentiable at x= 0 so f \in \mathcal{C}^{1}(E) but f is not in \mathcal{C}^{1}(E). A slightly more complicated example is f(x)= x^2sin(1/x) if x is not 0, f(0)= 0. It is fairly easy to show that f is continuous in any neighborhood of x= 0 and, in fact, that it has a derivative for all x, including x= 0, but the derivative function is not continuous at x= 0.
 
Thank you Mr. Hall,

I have a follow-up question.

Let say we have autonomous differential equation

x' = f(x) and that we have an open set E \subset \mathbb{R} and that f \in \mathcal{C}^1(E). We then assume that (I,x) is a solution to the diff-eqn also that x(t_1) = x(t_2) where t_1, t_2 \in I and that t_1 < t_2

Show for n = 1 that y is a constant solution.

I know from the definition of f that it has first derivatives on E and that E is open. And since from the definition of the solution to a diff.eqn that I \subset \mathbb{R}^n. Since I is considered to be a subset of R and that E is also a subset of \mathbb{R} for n = 1.

Then here is my questions:

1) Do I need to show that f also has derivatives on I?

2) I can't well assume that since f has derivatives on a subset of R then if f's domain consistets of both of R and E. Can this domain be seen a as closed set? And I then assume by Rollo's theorem that there exists some value on f's domain, t_3. Where if x is a solution of f, and then x'(t_1) = f(x(t_1)) = x'(t_2) = 0 and then x'(t_3) = f(x(t_3))= 0 and than the solution x is constant?
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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