Explanation Given : If a > 0 , x 3 + a x − 1 = 0 The given equation is x 3 + a x − 1 = 0 . Let, f ( x ) = x 3 + a x − 1 f ( 0 ) = 0 3 + a ( 0 ) − 1 = − 1 f ( 1 ) = 1 3 + a − 1 = a > 0 The polynomial is a continuous function. The sign change from negative to positive. It will show have at least one real zero. By using intermediate theorem. ∵ − 1 < 0 < a f ( 0 ) < f ( x 1 ) < f ( 1 ) It has at least one root between ( 0 , 1 ) . Let it has two real roots, x 1 , x 2

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 4.2, Problem 57PS

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Chapter 4.2, Problem 56PS

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Chapter 4 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

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The equation x 3 + a x − 1 = 0 has exactly one real solution.

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Chapter 4 Solutions