Answer
None
Work Step by Step
$$\eqalign{
& g\left( t \right) = {t^5} + 5{t^3} + 50t \cr
& {\text{Differentiate the given function}} \cr
& g'\left( t \right) = \frac{d}{{dt}}\left[ {{t^5} + 5{t^3} + 50t} \right] \cr
& g'\left( t \right) = \frac{d}{{dt}}\left[ {{t^5}} \right] + \frac{d}{{dt}}\left[ {5{t^3}} \right] + \frac{d}{{dt}}\left[ {50t} \right] \cr
& g'\left( v \right) = \frac{d}{{dt}}\left[ {{t^5}} \right] + 5\frac{d}{{dt}}\left[ {{t^3}} \right] + 50\frac{d}{{dt}}\left[ t \right] \cr
& {\text{Recall that }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}},{\text{ and }}\frac{d}{{dx}}\left[ x \right] = 1 \cr
& g'\left( t \right) = 5{t^4} + 5\left( {3{t^2}} \right) + 50\left( 1 \right) \cr
& g'\left( t \right) = 5{t^4} + 15{t^2} + 50 \cr
& {\text{The domain of }}g'\left( t \right){\text{ is }}\left( { - \infty ,\infty } \right),{\text{ }}g'\left( v \right){\text{ exists for all }}t,{\text{ then}} \cr
& {\text{the critical numbers occur when }}g'\left( t \right) = 0 \cr
& 5{t^4} + 15{t^2} + 50 = 0 \cr
& 5{\left( {{t^2}} \right)^2} + 15\left( {{t^2}} \right) + 50 = 0 \cr
& {\text{Solve by the quadratic formula}} \cr
& {t^2} = \frac{{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4\left( 5 \right)\left( {50} \right)} }}{{2\left( 5 \right)}} \cr
& {t^2} = \frac{{ - 15 \pm \sqrt { - 775} }}{{2\left( 5 \right)}} \cr
& {\text{The equation has no real solutions}}{\text{, then there are no critical}} \cr
& {\text{points.}} \cr} $$
