Answer
$v = - 2{\text{ and }}v = 2$
Work Step by Step
$$\eqalign{
& g\left( v \right) = {v^3} - 12v + 4 \cr
& {\text{Differentiate the given function}} \cr
& g'\left( v \right) = \frac{d}{{dv}}\left[ {{v^3} - 12v + 4} \right] \cr
& g'\left( v \right) = \frac{d}{{dv}}\left[ {{v^3}} \right] - \frac{d}{{dv}}\left[ {12v} \right] + \frac{d}{{dv}}\left[ 4 \right] \cr
& {\text{Recall that }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}},{\text{ and }}\frac{d}{{dx}}\left[ c \right] = 0 \cr
& g'\left( v \right) = 3{v^2} - 12 + 0 \cr
& g'\left( v \right) = 3{v^2} - 12 \cr
& {\text{The domain of }}g'\left( v \right){\text{ is }}\left( { - \infty ,\infty } \right),{\text{ }}g'\left( v \right){\text{ exists for all }}v,{\text{ then}} \cr
& {\text{the critical numbers occur when }}g'\left( v \right) = 0 \cr
& 3{v^2} - 12 = 0 \cr
& {\text{Factoring}} \cr
& 3\left( {{v^2} - 4} \right) = 0 \cr
& 3\left( {v + 2} \right)\left( {v - 2} \right) = 0 \cr
& {\text{Zero - factor property}} \cr
& v + 2 = 0,{\text{ }}v - 2 = 0 \cr
& v = - 2{\text{ and }}v = 2 \cr} $$
