## Calculus 10th Edition

Published by Brooks Cole

# Chapter 2 - Differentiation - 2.3 Exercises: 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.

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