Answer
(a) $ 1.28 \ \text{cm}/\text{year }$
(b) $t\approx 2.33\ \text{ years}$
(c) $ 32 \mathrm{cm}$
Work Step by Step
Given $$L(t)=32\left(1-e^{-0.37 t}\right)$$
(a) Since
$$ L'(t)=32\left(0.37e^{-0.37 t}\right)= 11.84e^{-0.37 t}$$
at $t=6$, we get
\begin{align*}
L'(6)&= 11.84e^{-0.37 (6)}\\
&= 1.28 \ \text{cm}/\text{year }
\end{align*}
(b) Since
\begin{align*}
L'(t)&=11.84e^{-0.37 t}\\
5&= 11.84e^{-0.37 t}\\
\ln (0.42229)&= -0.37t
\end{align*}
Then $$t\approx 2.33\ \text{ years}$$
(c) Since \begin{align*}
L&=\lim _{t \rightarrow \infty}\left(32-32 e^{-0.37 t}\right)\\
&=32-32 \lim _{t \rightarrow \infty} e^{-0.37 t}\\
&=32-0\\
&=32 \mathrm{cm}
\end{align*}
