Chapter 4.6, Problem 108E

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# PUTNAM EXAM CHALLENGELet f(x) be defined for a ≤ x ≤ b . Assuming appropriate properties of continuity and derivability, prove for a < x < b that f ( a ) − f ( a ) x − a − f ( b ) − f ( a ) b − a x − b = 1 2 f " ( ε ) ,where ε is some number between a and b.This problem was composed by the Committee on the Putnam Prize Competition.© The Mathematical Association of America. All rights reserved.

To determine

To prove: For a<x<b, f(x)f(a)xaf(b)f(a)baxb=12f''(ε) where ε lies between a and b.

Explanation

Given:

The function f is continuous in the interval [a,b] and differentiable on the interval (a,b).

Formula used:

Rolleâ€™s Theorem states that for any function g(x) that is differentiable in the interval (a,b) and continuous in the interval [a,b], if g(a)=g(b), then there exists c in the interval (a,b) such that f'(c)=0.

Proof:

Let p=f(x)âˆ’f(a)xâˆ’aâˆ’f(b)âˆ’f(a)bâˆ’axâˆ’b.

This can be further simplified to obtain:

p(xâˆ’b)=f(x)âˆ’f(a)xâˆ’aâˆ’f(b)âˆ’f(a)bâˆ’ap(xâˆ’b)(xâˆ’a)=f(x)âˆ’f(a)âˆ’f(b)âˆ’f(a)bâˆ’a(xâˆ’a)f(x)=f(a)+f(b)âˆ’f(a)bâˆ’a(xâˆ’a)+p(xâˆ’b)(xâˆ’a)

Now define a function q(t) as:

q(t)=f(t)âˆ’f(a)âˆ’f(b)âˆ’f(a)bâˆ’a(tâˆ’a)âˆ’p(tâˆ’b)(tâˆ’a)

Now,

q(a)=f(a)âˆ’f(a)<

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