Answer
$x = - \frac{1}{6}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 3{x^2} + x - 2 \cr
& {\text{Differentiate the given function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {3{x^2} + x - 2} \right] \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {3{x^2}} \right] + \frac{d}{{dx}}\left[ x \right] - \frac{d}{{dx}}\left[ 2 \right] \cr
& {\text{Recall that }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}}{\text{ and }}\frac{d}{{dx}}\left[ c \right] = 0 \cr
& f'\left( x \right) = 6x + 1 - 0 \cr
& f'\left( x \right) = 6x + 1 \cr
& {\text{The domain of }}f'\left( x \right){\text{ is }}\left( { - \infty ,\infty } \right),{\text{ }}f'\left( x \right){\text{ exists for all }}x,{\text{ then}} \cr
& {\text{the critical numbers occur when }}f'\left( x \right) = 0 \cr
& 6x + 1 = 0 \cr
& {\text{Solve for }}x \cr
& 6x = - 1 \cr
& x = - \frac{1}{6} \cr} $$
