Appl of IVT - ≤ 0 ≤ f(0 – 0 = g(0(as 0 ≤ f x ≤ 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem Proof. Let g : [0, 1] R be the function defined by g ( x ) = f ( x ) x . Note that (i) g is continuous (as f is continuous) on [0, 1] and (ii) g (1) = f (1) – 1
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ≤ 0 ≤ f (0) – 0 = g (0) (as 0 ≤ f ( x ) ≤ 1). Thus, by IVT , there exists c in [0, 1] s.t. g ( c ) = 0, i.e., f ( c ) = c . ...
View Full Document

This note was uploaded on 05/10/2011 for the course MATH 1505 taught by Professor Yap during the Winter '11 term at National University of Singapore.

Ask a homework question - tutors are online