Is a graph Continuous and differentiable at a given point

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Homework Help Overview

The discussion revolves around the continuity and differentiability of a piecewise function defined as f(x)={(2x-1)/|2x-1| for x ≠ 1/2, 0 for x = 1/2. Participants are examining the behavior of the function at the point x = 1/2, particularly focusing on whether the function is continuous and differentiable at that point.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to visualize the function by describing its graph and identifying points of interest. There are discussions about the nature of the graph, including its continuity and the implications for differentiability. Questions are raised about how to determine continuity at x = 1/2 and the relationship between continuity and differentiability.

Discussion Status

The discussion is active, with participants sharing their interpretations of the graph and questioning the continuity and differentiability of the function at the specified point. Some participants have provided guidance on how to approach the problem, while others are clarifying concepts related to differentiability and continuity.

Contextual Notes

Participants are navigating through the definitions of continuity and differentiability, with some confusion about the terminology used to describe these concepts. There is an emphasis on ensuring accurate understanding of the relationship between the two properties.

betsinda
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Homework Statement



F
f(x)={(2x-1)/Absolute value(2x-1) x cannot equal (1/2)
{ 0 x = (1/2)

a) is f continuous at X = (1/2) explain
b) is f differentiable at x = (1/2) explain

Homework Equations



I have made the graph and x is a point at 1/2 but there is a jump. I have no idea how to start this.



The Attempt at a Solution

 
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Drawing a graph is a good start. Can you describe what the graph looks like?
 
the line from the left approaches (1/2) at +1 and restarts at (1/2) at -1. There is a single point at (0,1/2)
 
sorry it continues to the right at -1
 
betsinda said:
the line from the left approaches (1/2) at +1 and restarts at (1/2) at -1. There is a single point at (0,1/2)

No, none of this is right. The graph of this function is in three parts--two horizontal lines and a single point.

What is f(-1)? f(0)? f(1/2)? f(1)? f(2)? Plotting these points should give you an idea of what the graph of the function looks like.
 
okay, I have fixed the graph. I have a line going from (1/2, 1) (1,1)(2,1) ect and a line going from (1/2,-1)(0,-1)(1,-1) ect, and a point at (1/2,0).

How do i determine if this is continuous at 1/2?
 
One quibble. The line doesn't contain the point (1/2, -1). Does the graph look continuous at x = 1/2?
 
I would say that the graph is not continuous at x=1/2 as that point does not connect to any other point on the graph.
 
  • #10
would it then be correct to say that x=1/2 is not differentiable as it is not continuous at that point?
 
  • #11
based on the fact that well every function is not differentiable, very function that is differentiable is continous. Or am I misunderstanding the concept?
 
  • #12
betsinda said:
would it then be correct to say that x=1/2 is not differentiable as it is not continuous at that point?
No. It doesn't make any sense to talk about a point or an x value being differentiable. You can say, though, that a function is continuous or differentiable at a point or at some x value.
 
  • #13
betsinda said:
based on the fact that well every function is not differentiable, very function that is differentiable is continous. Or am I misunderstanding the concept?
That is correct, so you are not misunderstanding the concept. I have made a couple of edits to what you wrote:
While not every function is differentiable, every function that is differentiable is continous.
 
  • #14
B) is f differentiable at x = (1/2) explain

So I could say that function is not differentiable at x=1/2 as the function is not not continuous?

Sorry for being a little slow I'm just trying to wrap my head around the concept. Thanks
 

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