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: 61


At the point $(2,4)$, $f(x)$ is parallel with the given line.

Work Step by Step

$$f(x)=3x^2-4x$$ 1) Find $f'(x)$: $$f'(x)=(3x^2-4x)'=6x-4$$ 2) The slope of the tangent line to $f(x)$ at $x=a$ is $f'(a)=6a-4$. We know that parallel lines have the same slope. Therefore, all the tangent lines to $f(x)$ parallel to $y=8x+5$ must have $f'(a)=8$. In other words, in order to find the required points, we now have to find $a$ such that $f'(a)=6a-4=8$: $$6a-4=8$$ $$6a=12$$ $$a=2$$ For $a=2$: $$f(a)=3\times2^2-4\times2=3\times4-8=12-8=4$$ Therefore, at the point $(2,4)$, $f(x)$ is parallel with the given line.
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