Explanation Given information: p is a function that is defined for all x > 0 and satisfies p ( a b ) = p ( b ) − p ( a ) . Function f is defined by f ( x ) = p ( 2 − x 2 + x ) . Concepts used: 1. ) A function f is even if f ( − x ) = f ( x ) . 2. ) A function f is odd if f ( − x ) = − f ( x ) . Calculation: As per the given information, p is a function that is defined for all x > 0 and satisfies p ( a b ) = p ( b ) − p ( a ) . Function f is defined by, f ( x ) = p ( 2 − x 2 + x ) … (1) Since, p ( a b ) = p ( b ) − p ( a ) ...(2) Therefore, by equation (2), f ( x ) = p ( 2 − x 2 + x ) = p ( 2 + x ) − p ( 2 − x )

4th Edition

ISBN: 9781319545956

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 1.1, Problem 69E

To determine

To prove:

The function f(x)=p(2−x2+x) is an odd function.

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Chapter 1.1, Problem 68E

Chapter 1.1, Problem 70E

Chapter 1 Solutions

CALCULUS W/ACHIEVE PLATFORM

Ch. 1.1 - Prob. 1PQ Ch. 1.1 - Prob. 2PQ Ch. 1.1 - Prob. 3PQ Ch. 1.1 - Prob. 4PQ Ch. 1.1 - Prob. 5PQ Ch. 1.1 - Prob. 6PQ Ch. 1.1 - Prob. 7PQ Ch. 1.1 - Prob. 8PQ Ch. 1.1 - Prob. 9PQ Ch. 1.1 - Prob. 10PQ

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - 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Prob. 56E Ch. 1.5 - Prob. 57E Ch. 1.5 - Prob. 58E Ch. 1.5 - Prob. 59E Ch. 1.5 - Prob. 60E Ch. 1.6 - Prob. 1PQ Ch. 1.6 - Prob. 2PQ Ch. 1.6 - Prob. 3PQ Ch. 1.6 - Prob. 4PQ Ch. 1.6 - Prob. 5PQ Ch. 1.6 - Prob. 6PQ Ch. 1.6 - Prob. 7PQ Ch. 1.6 - Prob. 8PQ Ch. 1.6 - Prob. 9PQ Ch. 1.6 - Prob. 1E Ch. 1.6 - Prob. 2E Ch. 1.6 - Prob. 3E Ch. 1.6 - Prob. 4E Ch. 1.6 - Prob. 5E Ch. 1.6 - Prob. 6E Ch. 1.6 - Prob. 7E Ch. 1.6 - Prob. 8E Ch. 1.6 - Prob. 9E Ch. 1.6 - Prob. 10E Ch. 1.6 - Prob. 11E Ch. 1.6 - Prob. 12E Ch. 1.6 - Prob. 13E Ch. 1.6 - Prob. 14E Ch. 1.6 - Prob. 15E Ch. 1.6 - Prob. 16E Ch. 1.6 - Prob. 17E Ch. 1.6 - Prob. 18E Ch. 1.6 - Prob. 19E Ch. 1.6 - Prob. 20E Ch. 1.6 - Prob. 21E Ch. 1.6 - Prob. 22E Ch. 1.6 - Prob. 23E Ch. 1.6 - Prob. 24E Ch. 1.6 - Prob. 25E Ch. 1.6 - Prob. 26E Ch. 1.6 - Prob. 27E Ch. 1.6 - Prob. 28E Ch. 1.6 - Prob. 29E Ch. 1.6 - Prob. 30E Ch. 1.6 - Prob. 31E Ch. 1.6 - Prob. 32E Ch. 1.6 - Prob. 33E Ch. 1.6 - Prob. 34E Ch. 1.6 - Prob. 35E Ch. 1.6 - Prob. 36E Ch. 1.6 - Prob. 37E Ch. 1.6 - 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Finding Local Maxima and Minima by Differentiation; Author: Professor Dave Explains;https://www.youtube.com/watch?v=pvLj1s7SOtk;License: Standard YouTube License, CC-BY

To prove: The function f ( x ) = p ( 2 − x 2 + x ) is an odd function.

Solution Summary: The author explains that the function f(x)=p(2-x2+x+) is an odd function.

Concept explainers

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To prove:

The function f(x)=p(2−x2+x) is an odd function.

Want to see the full answer?

Chapter 1 Solutions

To prove: The function f ( x ) = p ( 2 − x 2 + x ) is an odd function.

Solution Summary: The author explains that the function f(x)=p(2-x2+x+) is an odd function.

Concept explainers

Videos

To prove: The function f(x)=p(2−x2+x) is an odd function.

Want to see the full answer?

Chapter 1 Solutions

To prove:

The function f(x)=p(2−x2+x) is an odd function.