#### Answer

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

#### 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*}