Inflection point of non continuous or non differentiable function

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Homework Help Overview

The discussion revolves around identifying inflection points for three specific functions, including a piecewise function, a rational function, and an absolute value function. The participants explore the implications of continuity and differentiability on the existence of inflection points.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to analyze the concavity of the functions and questions whether continuity is a necessary condition for inflection points. They specifically inquire about the implications of non-differentiability and points not being in the domain.

Discussion Status

Some participants suggest that the definition of an inflection point may vary based on the text being referenced, indicating that the first two functions may not qualify due to continuity issues, while the third function might be considered valid due to a change in concavity.

Contextual Notes

There is an ongoing discussion about the definitions of inflection points and the role of continuity and differentiability in determining their existence. The original poster's functions present specific challenges related to these concepts.

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Homework Statement


three functions:
[tex]y=\begin{cases}\arctan \frac{1}{x}\ x\neq0\\ 0\ x=0\end{cases}[/tex]
[tex]y=\frac{1}{x}, y=|x^2-1|[/tex] and what about inflection point?

The Attempt at a Solution


first function is concave on left of 0, convex on right, so from definition it should be inflection point, but its not continuous in this point, a function need to be continuous in this place or not?
in 2, [tex]x=0[/tex] should be inflection point, but its not in the domain, so is there inflection point?
in 3, function is continuous in [tex]x=1[/tex] but not differentiable, is there inflection point or not?
 
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player1_1_1 said:

Homework Statement


three functions:
[tex]y=\begin{cases}\arctan \frac{1}{x}\ x\neq0\\ 0\ x=0\end{cases}[/tex]
[tex]y=\frac{1}{x}, y=|x^2-1|[/tex] and what about inflection point?

The Attempt at a Solution


first function is concave on left of 0, convex on right, so from definition it should be inflection point, but its not continuous in this point, a function need to be continuous in this place or not?
in 2, [tex]x=0[/tex] should be inflection point, but its not in the domain, so is there inflection point?
in 3, function is continuous in [tex]x=1[/tex] but not differentiable, is there inflection point or not?

It probably depends on the definition your text gives. Most say it must be a point on the graph where the concavity changes. That would rule out the first two. I would say the third qualifies because of the change in concavity at the point. But your mileage may vary.
 
thx!
 

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