inflection point of non continuous or non differentiable function


by player1_1_1
Tags: continuous, differentiable, function, inflection, point
player1_1_1
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#1
Feb11-11, 04:29 AM
P: 118
1. The problem statement, all variables and given/known data
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?
3. 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
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#2
Feb11-11, 02:29 PM
P: 118
up,.
LCKurtz
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#3
Feb11-11, 02:42 PM
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Quote Quote by player1_1_1 View Post
1. The problem statement, all variables and given/known data
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?
3. 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.

player1_1_1
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#4
Feb12-11, 02:14 AM
P: 118

inflection point of non continuous or non differentiable function


thx!


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