Can You Prove f(100) Is Less Than 100 Given f(0) = 0 and a Specific Derivative?

  • Thread starter Thread starter flyingpig
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around proving that f(100) is less than 100, given the conditions f(0) = 0 and the derivative f'(x) = 1/(1 + e^(-f(x))). Participants are exploring the implications of these conditions in the context of calculus and analysis.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of integrals to express f(100) and consider the application of the Mean Value Theorem. There are attempts to establish bounds on the integral based on the properties of the derivative.

Discussion Status

The conversation is ongoing, with participants attempting to clarify the relationship between the integral representation of f(100) and its upper bound. Some guidance has been offered regarding the properties of the derivative, but no consensus has been reached on the proof itself.

Contextual Notes

Participants are working under the assumption that f(0) = 0 and are examining the behavior of the function and its derivative without additional information about f(x) beyond what is provided.

flyingpig
Messages
2,574
Reaction score
1

Homework Statement

For f(0) = 0, and that f'(x) = [tex]\frac{1}{1 + e^{-f(x)}}[/tex], prove that f(100) < 100

The Attempt at a Solution



I did

[tex]\f(100) = \int_{0}^{100} \frac{dx}{1 + e^{-f(x)}}[/tex]

Unfortunately, I got f(100) back...
 
Physics news on Phys.org
flyingpig said:
I did

[tex]\f(100) = \int_{0}^{100} \frac{dx}{1 + e^{-f(x)}}[/tex]

Unfortunately, I got f(100) back...

That's already ok. Can you prove now that the integral must be <100??

First, can you prove that

[tex]\frac{1}{1+e^{-f(x)}}\leq 1[/tex]
 
flyingpig said:

Homework Statement




For f(0) = 0, and that f'(x) = [tex]\frac{1}{1 + e^{-f(x)}}[/tex], prove that f(100) < 100


The Attempt at a Solution



I did

[tex]\f(100) = \int_{0}^{100} \frac{dx}{1 + e^{-f(x)}}[/tex]

Unfortunately, I got f(100) back...

For y > = 0 we have 1/(1+exp(-y)) = exp(y)/[1+exp(y)] <= 1.

RGV
 

Similar threads

Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
5K
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 48 ·
2
Replies
48
Views
7K