Lars Olsen proof of Darboux's Intermediate Value Theorem for Derivatives

Click For Summary
SUMMARY

Lars Olsen's proof of Darboux's Intermediate Value Theorem for Derivatives establishes that for differentiable functions, the value ##y## lies between the values of the functions ##f_a(a)## and ##f_a(b)##, or ##f_b(a)## and ##f_b(b)##. The proof hinges on the relationship ##f_a(b) = f_b(a)##, which is derived from the Mean Value Theorem. The discussion clarifies the conditions under which ##y## is bounded by the derivatives at points ##a## and ##b##, specifically ##f'(a) < y < f'(b)##. The participants emphasize the importance of understanding the connection between the endpoints of the graphs of the functions involved.

PREREQUISITES
  • Understanding of the Mean Value Theorem
  • Familiarity with differentiable functions
  • Knowledge of the concepts of limits and continuity
  • Basic grasp of derivative notation and properties
NEXT STEPS
  • Study the Mean Value Theorem in detail
  • Explore the implications of Darboux's Theorem in calculus
  • Investigate examples of differentiable functions that illustrate the theorem
  • Learn about the relationship between derivatives and function behavior
USEFUL FOR

Mathematicians, calculus students, educators, and anyone interested in the properties of derivatives and their applications in real analysis.

Hall
Messages
351
Reaction score
87
Here is Lars Olsen's proof. I'm having difficulty in understanding why ##y## will lie between ##f_a (a)## and ##f_a(b)##. Initially, we assumed that ##f'(a) \lt y \lt f'(b)##, but ##f_a(b)## doesn't equal to ##f'(b)##.
 
Last edited by a moderator:
Physics news on Phys.org
##y## will lie between ##f_a(a)## and ##f_b(b)##.
We also have by the construction of the functions ##f_a## and ##f_b## that ##f_a(b) = f_b(a)##.
Thus, ##y## will lie between ##f_a(a)## and ##f_a(b)## or between ##f_b(a)## and ##f_b(b)##.

What is it that you do not understand? Can you come up with a differentiable function ##f## for which the above is not true?
 
malawi_glenn said:
##y## will lie between ##f_a(a)## and ##f_b(b)##.
We also have by the construction of the functions ##f_a## and ##f_b## that ##f_a(b) = f_b(a)##.
Thus, ##y## will lie between ##f_a(a)## and ##f_a(b)## or between ##f_b(a)## and ##f_b(b)##.

What is it that you do not understand? Can you come up with a differentiable function ##f## for which the above is not true?
## f_a(a) = f'(a) \lt y \lt f'(b) = f_b(b)##, that is correct. But how ##f_a(b) = f'(b)##?
 
Hall said:
But how fa(b)=f′(b)?
where is that written/used?
In other words, where do you read ##f_a(b) = f_b(b)##?
 
malawi_glenn said:
where is that written/used?
In other words, where do you read ##f_a(b) = f_b(b)##?
I don't know, sir, but I cannot see simply how ##y## lies between ##f_a(a)## and ##f_a(b)##.
 
##f_a(b) = \frac{ f(a) - f(b)}{a-b}##, by Mean Value Theorem there is some ##c \in (a,b)## such that ##f_a(b) = f'(c)##. Why ##y## necessarily need to lie between ##f'(a) and f'(c)##? We assumed it to lie between ##f'(a)## and ##f'(b)##.
 
Hall said:
I don't know, sir, but I cannot see simply how ##y## lies between ##f_a(a)## and ##f_a(b)##.
You have two cases to consider.
As I wrote, which is in the article.:
We have by the construction of the functions ##f_a## and ##f_b## that ##f_a(b) = f_b(a)##.
Thus, ##y## will lie between ##f_a(a)## and ##f_a(b)## or between ##f_b(a)## and ##f_b(b)##.

1672245477224.png


Work it out with the function say ##f(x) = x^2## and let ##a=-1## and ##b=2## or something.
 
Last edited:
  • Like
Likes   Reactions: jedishrfu
Darboux.jpg
.

If ##y## lies between ##f'(a)## and ##f'(b)##, then it is surely to between the end-points of one of the graphs.

It was not very obvious for me to see, I was considering ##f_a(a)## and ##f_a(b)## individually and not in connection with that "or" ##f_b(a)## and ##f_b(a)##.
 
Are you okay with the proof now?
 
  • Like
Likes   Reactions: jedishrfu
  • #10
malawi_glenn said:
Are you okay with the proof now?
Yes.
 
  • Love
  • Like
Likes   Reactions: jedishrfu and malawi_glenn

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K