University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 118: 26


The graph of $f$ has horizontal tangents at $(1,-2)$ and $(-1,2)$.

Work Step by Step

$$f(x)=x^3-3x$$ Having a horizontal tangent means that having the slope, and the derivative, there equal $0$. In other words, we need to find $x$ such that $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=0\hspace{1cm}(1)$$ First, we examine $f(x+h)$: $f(x+h)=(x+h)^3-3(x+h)=x^3+3x^2h+3xh^2+h^3-3x-3h=(x^3-3x)+(3x^2h+3xh^2+h^3-3h)$ Therefore, $$\frac{f(x+h)-f(x)}{h}=\frac{3x^2h+3xh^2+h^3-3h}{h}=3x^2+3xh+h^2-3$$ Replacing the result into $(1)$: $$\lim_{h\to0}(3x^2+3xh+h^2-3)=0$$ $$3x^2+3x\times0+0^2-3=0$$ $$3x^2-3=0$$ $$x^2-1=0$$ $$x=\pm1$$ We have $f(1)=1^3-3\times1=1-3=-2$ and $f(-1)=(-1)^3-3\times(-1)=-1+3=2$ Therefore, the graph of $f$ has horizontal tangents at $(1,-2)$ 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.