1.8.6 AP calculus Exam continuity

Click For Summary

Discussion Overview

The discussion revolves around the concept of continuity in calculus, specifically examining the function $\dfrac{(2x+1)(x-2)}{x-2}$ and its relationship to the function $(2x+1)$. Participants explore the implications of continuity at a point, particularly at $x=2$, and the significance of removable discontinuities.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses dissatisfaction with the explanation of why $\dfrac{(2x+1)(x-2)}{x-2}$ is not the same as $(2x+1)$.
  • Another participant suggests that the concept of filling a hole in the graph relates to the continuity discussion.
  • A definition of continuity at a point $x=c$ is provided, outlining the conditions for a function to be continuous.
  • A later reply elaborates on the definition of continuity, emphasizing the importance of limits and the irrelevance of the function's value at the point of discontinuity.
  • The same participant argues that for $x \ne 2$, the two functions are equivalent, but highlights that the first function is undefined at $x=2$, creating a discontinuity.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the explanation of the functions' equivalence and the nature of continuity, indicating ongoing debate and differing perspectives on the topic.

Contextual Notes

The discussion highlights the nuances of continuity definitions and the implications of removable discontinuities, with some participants emphasizing different aspects of the definitions and their applications.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
View attachment 9447
well just by observation $x-2$ cancels out leaving $2x+1$ plugging in $2$ gives $2(2)+1 = 5$

or is there some more anointed way to do this.
 

Attachments

  • Capture.PNG
    Capture.PNG
    2.7 KB · Views: 145
Last edited:
Physics news on Phys.org
Not really a fan of this explanation.

I want to know why

$\dfrac{(2x+1)(x-2)}{x-2}$

is NOT EXACTLY THE SAME as

$(2x+1)$
 
I assume its about k

Always thought filling a hole was more rigorous
 
definition of continuity at $x=c$ ...

1. $f(c)$ exists

2. $\displaystyle \lim_{x \to c} f(x)$ exists

3. $\displaystyle \lim_{x \to c} f(x) = f(c)$
 
I shall personally anoint this:
The definition of "continuous" is "f is continuous at x= a if and only if both f(a) and \lim_{x\to a} f(x) exist and
\lim_{x\to a} f(x)= f(a)". Further
\lim_{x\to a} f(x)= L is defined by "given any \epsilon> 0, there exist \delta> 0 such that if 0< |x- a|< \delta then |f(x)- L|< \epsilon.

Note the "0< |x- a|" which perhaps should be given more emphasis in math classes. The value of \lim_{x\to a} f(x) depends entirely upon the values of f(x) for x close to x= a, but the value of f(a), or even whether it exists, is completely irrelevant to the limit!

For this problem, as long as x is not 2, x- 2 is not 0 and the "x- 2" terms in numerator and denominator cancel.

That is, for x\ne 2, f(x)= \frac{(2x+1)(x- 2)}{x- 2}= 2x+1.

tkhunny said:
Not really a fan of this explanation.

I want to know why

$\dfrac{(2x+1)(x-2)}{x-2}$

is NOT EXACTLY THE SAME as

$(2x+1)$
They are "NOT EXACTLY THE SAME" because the second has value 5 when x= 2 and the first is not defined there. The graph of y= 2x+ 1 is a straight line while the graph of y=\frac{(2x+1)(x- 2)}{x- 2} is that same straight line except for a hole at (2, 5).


 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K