Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1 - Page 108: 26


$(1,-2)$ and $(-1,2)$

Work Step by Step

Given that $f(x)=x^3-3x$, the function will have a horizontal tangent when the derivative is equal to 0, since the slope will indicate that there is no change in the y value of the tangent when x changes. $f'(x)$ is found using the power rules: $f'(x)=3x^2-3$ We let $f'(x)$ be equal to zero. $3x^2-3=0$ $x^2=1$ $x=+1$ or $x=-1$ $f(1)=-2$ and $f(-1)=2$ Thus, there exist two different maxima and/or minima on the function where both points have horizontal tangents. $(1,-2)$ and $(-1,2)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.