Answer
(a) $\lim\limits_{t \to \infty}C(t) = 0$
(b) $C'(t) = K(be^{-bt}-ae^{-at})$
(c) $C'(t) = 0~~$ when $~~t = \frac{ln(b)-ln(a)}{b-a}$
Work Step by Step
(a) $C(t) = K(e^{-at}-e^{-bt})$
$\lim\limits_{t \to \infty}C(t)$
$= \lim\limits_{t \to \infty}K(e^{-at}-e^{-bt})$
$= \lim\limits_{t \to \infty}K(\frac{1}{e^{at}}-\frac{1}{e^{bt}})$
$= K(0-0)$
$= 0$
(b) $C(t) = K(e^{-at}-e^{-bt})$
$C'(t) = K(-ae^{-at}+be^{-bt})$
$C'(t) = K(be^{-bt}-ae^{-at})$
(c) $C'(t) = K(be^{-bt}-ae^{-at}) = 0$
$K(be^{-bt}-ae^{-at}) = 0$
$be^{-bt}= ae^{-at}$
$ln(be^{-bt})= ln(ae^{-at})$
$ln(b)+ln(e^{-bt})= ln(a)+ln(e^{-at})$
$ln(b)-bt= ln(a)-at$
$bt-at = ln(b)-ln(a)$
$t(b-a) = ln(b)-ln(a)$
$t = \frac{ln(b)-ln(a)}{b-a}$
$C'(t) = 0~~$ when $~~t = \frac{ln(b)-ln(a)}{b-a}$
