Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises - Page 114: 30

Answer

a) $y=4x $; $y=-3x+18$ b) See graph

Work Step by Step

a) First, we have to differentiate the function $G(x) = 4x^{2}-x^{3}$, and we will get tangent line equation: $G^{'}(x) = 8x - 3x^{2}$ To get the gradient of the tangent line, we have to substitute each point with the points. For point $(2,8)$ $G^{'}(2) = 8(2) - 3(2)^{2} = 4$ For point $(3,9)$ $G^{'}(3) = 8(3) - 3(3)^{2} = -3$ Equation of the tangent line at point $(2,8)$: $y-8=4(x-2)$ $y=4x$ Equation of the tangent line at point $(3,9)$: $y-9=-3(x-3)$ $y= -3x+18$ b) Graph the function and the tangents in the two points.
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.