Imvt & LMVT: Existence of Point with f(x)=x

  • Context: Undergrad 
  • Thread starter Thread starter Ananya0107
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the existence of a point where a continuous function f defined on the interval [0,1] satisfies the equation f(x) = x. This is framed as a theoretical problem often encountered in homework contexts.

Discussion Character

  • Homework-related
  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant states the problem of finding a point where f(x) = x for a continuous function f on [0,1].
  • Another participant notes that this is a standard homework exercise and inquires about the approaches taken by others.
  • A participant proposes an argument involving the construction of a 1 by 1 square on the coordinate plane, suggesting that the graph of any continuous function must intersect the diagonal line y = x.
  • Another participant suggests applying the Mean Value Theorem (MVT), stating that there exists a point c where the derivative f'(c) equals the slope of the line connecting the endpoints of the function over the interval, which is 1.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the methods or arguments presented. Multiple viewpoints and approaches are discussed without resolution.

Contextual Notes

The discussion includes various assumptions about the properties of continuous functions and the application of the Mean Value Theorem, which may depend on specific conditions not fully explored in the posts.

Who May Find This Useful

This discussion may be useful for students studying calculus, particularly those interested in fixed-point theorems and the application of the Mean Value Theorem.

Ananya0107
Messages
20
Reaction score
2
Let f:[0,1]→[0,1] be a continuous function . Show that there exists a point such that f(x) = x
 
Physics news on Phys.org
That's a fairly standard homework exercise. What have you tried?
 
Tell me if this argument is right... I constructed a 1by 1 square on the coordinate plane vertices (0,0) ,(0,1),(1,0),(1,1). Then I said that to graph any continuous function , the graph must touch or cut the square's diagonal , that is y=x.
 
Apply MVT.
According to MVT, there exists a point c such that, f'(c)={f(1)-f(0)}/(1-0)=1-0=1
Now dy/dx=1. So you can find y.
now substitute a point in the domain to get the value of integration constant.
 

Similar threads

Undergrad Riemann Integral: Does g(x)=f(x) Almost Everywhere?
  • · Replies 28 ·
Replies
28
Views
7K
Undergrad ##f(2x)=f^2(x)-2f(x)-1/2## then find ##f(3)##
  • · Replies 17 ·
Replies
17
Views
3K
Graduate How to Find Critical Points of function f(x,y,z)
  • · Replies 9 ·
Replies
9
Views
3K
MHB Implicit function theorem for f(x,y) = x^2+y^2-1
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Proving that convexity implies second order derivative being positive
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Why prove Taylor's theorem with p(x)=q(x)−((b−a)^n/(b−x)^n)q(a)?
  • · Replies 9 ·
Replies
9
Views
3K
Undergrad Continuity of Mean Value Theorem
  • · Replies 12 ·
Replies
12
Views
3K
MHB Solve Image of f(x)=-2(x+1)/(x-1)^2 in x from [0,1)U(1,3]
  • · Replies 2 ·
Replies
2
Views
1K
Undergrad Find f(z) given f(x, y) = u(x, y) + iv(x,y)
  • · Replies 8 ·
Replies
8
Views
1K
Undergrad Should the function f to be continuous for applying MVTI or not?
  • · Replies 4 ·
Replies
4
Views
2K