University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises - Page 136: 62


At 2 points $(-1,-5/6)$ and $(4,-5/3)$, $g(x)$ is parallel with the given line.

Work Step by Step

$$g(x)=\frac{1}{3}x^3-\frac{3}{2}x^2+1$$ 1) Find $g'(x)$: $$g'(x)=(\frac{1}{3}x^3-\frac{3}{2}x^2+1)'=x^2-3x$$ 2) The slope of the tangent line to $g(x)$ at $x=a$ is $g'(a)=a^2-3a$. Now we need to rewrite the equation of the mentioned line $$8x-2y=1$$ $$2y=8x-1$$ $$y=4x-\frac{1}{2}$$ We know that parallel lines have the same slope. Therefore, all the tangent lines to $g(x)$ parallel to $y=4x-1/2$ must have $g'(a)=4$. In other words, in order to find the required points, we now have to find $a$ such that $g'(a)=a^2-3a=4$: $$a^2-3a=4$$ $$a^2-3a-4=0$$ $$(a+1)(a-4)=0$$ $$a=-1\hspace{1cm}\text{or}\hspace{1cm}a=4$$ - For $a=-1$: $$g(a)=\frac{1}{3}\times(-1)^3-\frac{3}{2}\times(-1)^2+1=-\frac{1}{3}-\frac{3}{2}+1=-\frac{5}{6}$$ - For $a=4$: $$g(a)=\frac{1}{3}\times(4)^3-\frac{3}{2}\times(4)^2+1=\frac{64}{3}-24+1=-\frac{5}{3}$$ Therefore, at 2 points $(-1,-5/6)$ and $(4,-5/3)$, $g(x)$ is parallel with the given line.
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.