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

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

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*}