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: 70

Answer

The equation of the tangent is $ y=\frac{x}{2}+3$.

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=27; u'(x)=0$ $v(x)=x^2+9; v'(x)=2x$ $f'(x)=\frac{(0)(x^2+9)-(2x)(27)}{(x^2+9)^2}$ $=-\frac{54x}{(x^2+9)^2}$ $f'(-3)=-\frac{54(-3)}{((-3)^2+9)^2}=\frac{1}{2}$ Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-\frac{3}{2})=\frac{1}{2}(x+3)\rightarrow y=\frac{x}{2}+3 $.
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