Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - Chapter Review Exercises - Page 221: 10


$$L(h) =16-8h$$

Work Step by Step

Given $$V(h)=4 h(2-h)(4-2 h)=8h^3-32h^2+32h, \quad a=1$$ Since \begin{align*} V^{\prime}(h)&=24h^2-64h+32\\ V^{\prime}(1)&= -8 \end{align*} Then the linear approximation is given by \begin{align*} L(h)&=V^{\prime}(a)(h-a)+V(a)\\ &= -8(h-1)+ 8\\ &=16-8h \end{align*}
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