Answer
None
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2{x^3} + {x^2} + 8x \cr
& {\text{Differentiate the given function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^3} + {x^2} + 8x} \right] \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^3}} \right] + \frac{d}{{dx}}\left[ {{x^2}} \right] + \frac{d}{{dx}}\left[ {8x} \right] \cr
& f'\left( x \right) = 2\frac{d}{{dx}}\left[ {{x^3}} \right] + \frac{d}{{dx}}\left[ {{x^2}} \right] + 8\frac{d}{{dx}}\left[ x \right] \cr
& {\text{Recall that }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr
& f'\left( x \right) = 2\left( {3{x^2}} \right) + 2x + 8\left( {{x^0}} \right) \cr
& f'\left( x \right) = 6{x^2} + 2x + 8 \cr
& {\text{The domain of }}f'\left( x \right){\text{ is }}\left( { - \infty ,\infty } \right),{\text{ }}f'\left( x \right){\text{ exists for all }}x,{\text{ then}} \cr
& {\text{the critical numbers occur when }}f'\left( x \right) = 0 \cr
& 6{x^2} + 2x + 8 = 0 \cr
& {\text{Divide both sides by 2}} \cr
& 3{x^2} + x + 4 = 0 \cr
& {\text{Solve by the quadratic formula}} \cr
& x = \frac{{ - \left( 1 \right) \pm \sqrt {{{\left( 1 \right)}^2} - 4\left( 3 \right)\left( 4 \right)} }}{{2\left( 3 \right)}} \cr
& x = \frac{{ - 1 \pm \sqrt { - 47} }}{6} \cr
& {\text{The equation has no real solutions}}{\text{, then there are no critical}} \cr
& {\text{points.}} \cr
& {\text{}} \cr} $$
