Answer
$t = - \frac{1}{4}$
Work Step by Step
$$\eqalign{
& p\left( t \right) = t{e^{4t}} \cr
& {\text{Differentiate with respect to }}t \cr
& q\left( t \right) = \frac{d}{{dt}}\left[ {t{e^{4t}}} \right] \cr
& {\text{Using the product rule for derivatives}} \cr
& q\left( t \right) = t\frac{d}{{dt}}\left[ {{e^{4t}}} \right] + {e^{4t}}\frac{d}{{dt}}\left[ t \right] \cr
& {\text{Computing derivatives and simplify}} \cr
& q'\left( t \right) = t\left( {4{e^{4t}}} \right) + {e^{4t}}\left( 1 \right) \cr
& q'\left( t \right) = 4t{e^{4t}} + {e^{4t}} \cr
& q'\left( t \right) = \left( {4t + 1} \right){e^{4t}} \cr
& {\text{The derivative }}q'\left( t \right){\text{ exists for all real number }}t,{\text{ then}} \cr
& {\text{the critical numbers occur when }}q'\left( t \right) = 0 \cr
& \left( {4t + 1} \right){e^{4t}} = 0,{\text{ }}{e^{4t}} > 0{\text{ for all real numbers}} \cr
& 4t + 1 = 0 \cr
& t = - \frac{1}{4} \cr} $$
