Determining domain for C^1 function

In summary, determining the domain for a \( C^1 \) function involves identifying the set of real numbers where the function is continuously differentiable. This requires ensuring that both the function and its first derivative exist and are continuous across the specified interval. The domain can be affected by the presence of discontinuities, singularities, or restrictions imposed by the nature of the function itself, such as square roots or logarithms that require positive arguments. Ultimately, careful analysis of these factors leads to the correct identification of the domain.
  • #1
lys04
113
4
Homework Statement
Picture
Relevant Equations
Partial derivatives
The ####x partial derivative is equal to $$L \frac{4x}{5(x^{2}+y^{2})^{\frac{-3}{5}}}$$ and the partial for ##y## is $$L \frac{4y}{5(x^{2}+y^{2})^{\frac{-3}{5}}}$$
Using the limit definition of partial derivatives I got the partial wrt ##x## is $$L \frac{h^{\frac{4}{5}}}{h}$$ which doesn’t exist as ##h## goes to ##0##. Similar argument for partial wrt ##y##. This means that ##f## isn’t ##C^1## at the origin, right?

At every other point the partial derivatives exist and is continuous because it’s a composition of a polynomial of two variables and ##x^2/5##, so ##f## is ##C^1## at all points except the origin.

Is the reasoning correct?
IMG_0425.jpeg
 
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  • #2
I don’t think latex is working, not sure what’s wrong with it sorry
 
  • #3
lys04 said:
I don’t think latex is working, not sure what’s wrong with it sorry
One problem is that { and } don't match.
 
  • #4
Works very well with the additional }:

$$L \frac{4x}{5(x^{2}+y^{2})^{\frac{-3}{5}}}$$
 
  • #5
[itex]\sqrt[2n+1]{x}[/itex] is not differentiable at [itex]x = 0[/itex] for [itex]n \geq 1[/itex]. This follows from the fact that [itex]x^{2n+1}[/itex] has a point of inflection here.
 

FAQ: Determining domain for C^1 function

What is a C1 function?

A C1 function is a function that is continuously differentiable. This means that the function itself is continuous, and its first derivative also exists and is continuous across its domain. In mathematical terms, if a function f(x) is C1, then both f(x) and f'(x) are continuous functions.

How do I determine the domain of a C1 function?

To determine the domain of a C1 function, you need to identify all the values of the independent variable (usually x) for which the function is defined and both the function and its first derivative are continuous. This often involves finding where the function is undefined, such as points of division by zero or square roots of negative numbers, and ensuring that the first derivative does not have discontinuities.

Are there any specific types of functions that are always C1?

Yes, polynomial functions, trigonometric functions, exponential functions, and logarithmic functions are typically C1 on their domains. However, it is important to check the continuity of the first derivative to ensure the function remains C1 across the entire domain, particularly for piecewise functions or those involving absolute values.

What role does continuity play in determining the domain of a C1 function?

Continuity is crucial in determining the domain of a C1 function because both the function and its first derivative must be continuous. If there are points where the function or its derivative is not continuous, those points must be excluded from the domain. Thus, continuity directly influences the set of values for which the function is C1.

Can a function have a domain that is a union of intervals and still be C1?

Yes, a function can have a domain that is a union of intervals and still be C1. For example, a piecewise function defined on multiple intervals can be C1 if it is continuous across the boundaries of those intervals and its first derivative remains continuous as well. It is important to verify the continuity of both the function and its derivative at the boundaries of the intervals.

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