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A continuous function ƒ : [0, 1] → R satisfies ƒ(0) f(1). Show that for each integer n ≥ 1 there exits x such that f(x+(1/n)) = f(x). Is the same statement true for numbers other than 1/n? =

A continuous function ƒ : [0, 1] → R satisfies ƒ(0) f(1). Show that for each integer n ≥ 1 there exits x such that f(x+(1/n)) = f(x). Is the same statement true for numbers other than 1/n? =

College Algebra
College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter3: Functions
Section3.3: Rates Of Change And Behavior Of Graphs
Problem 2SE: If a functionfis increasing on (a,b) and decreasing on (b,c) , then what can be said about the local...
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A continuous function f [0, 1] → R satisfies f(0) = f(1). Show that for : each integer n ≥ 1 there exits x such that f(x+(1/n)) = f(x). Is the same statement true for numbers other than 1/n?
Transcribed Image Text:A continuous function f [0, 1] → R satisfies f(0) = f(1). Show that for : each integer n ≥ 1 there exits x such that f(x+(1/n)) = f(x). Is the same statement true for numbers other than 1/n?
Expert Solution
Step 1: Proof

Let fix δ=1n, n∈ℕ

Consider the continuous function gnx:0,1-δ→ℝ; gnx:=fx+1n-fx and suppose that gnx≠0 on I0:=0,1

Then by intermediate value theorem, gnx does not changes sign in I0.

Without losing of generality if gnx>0, then adding the n inequalities

gn1-1n=f1-f1-1n>0;gn1-2n=f1-1n-f1-2n>0;                             ⋯gn0=f1n-f0>0

yields f1>f0 which is a contradiction of our hypothesis that is f1=f0

Thus there exist a point x satisfying fx+1n=fx

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