Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.3 Exercises - Page 126: 63

Answer

The equation of the tangent is $y=-3x-1$.

Work Step by Step

Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=x^3+4x-1 ;u’(x)=3x^2+4 $. $v(x)=x-2 ;v’(x)=1 $. $f'(x)=(3x^2+4)(x-2)+(x^3+4x-1)$. $f'(1)=(3(1^2)+4)(1-2)+(1^3+4(1)-1)=-3$ Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y+4)=-3x+3 \rightarrow y=-3x-1$ A graphing calculator and a computer algebra system have been used to confirm these results.
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.