Answer
$${\text{Relative minimum at }}\left( {3, - 25} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^4} - 4{x^3} + 2 \cr
& {\text{*Calculate the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^4} - 4{x^3} + 2} \right] \cr
& f'\left( x \right) = 4{x^3} - 12{x^2} \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& 4{x^3} - 12{x^2} = 0 \cr
& {\text{Factoring}} \cr
& 4{x^2}\left( {x - 3} \right) \cr
& x = 0,{\text{ }}x = 3 \cr
& \cr
& *{\text{Calculate the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {4{x^3} - 12{x^2}} \right] \cr
& f''\left( x \right) = 12{x^2} - 24x \cr
& {\text{Evaluate the second derivative at }}x = 0{\text{ and }}x = 3 \cr
& *f''\left( 0 \right) = 12{\left( 0 \right)^2} - 24\left( 0 \right) = 0 \cr
& {\text{Then, the second derivative test fails}}\left( {{\text{Theorem 3}}{\text{.9}}} \right),{\text{ so we can}} \cr
& {\text{use the first derivative test at }}x = 0 \cr
& \left( { - \infty ,0} \right){\text{ and }}\left( {0,3} \right) \cr
& {\text{For the test values: }}x = - 1{\text{ and }}x = 1 \cr
& f'\left( { - 1} \right) = 4{\left( { - 1} \right)^3} - 12{\left( { - 1} \right)^2} = - 16 \cr
& f'\left( 1 \right) = 4{\left( 1 \right)^3} - 12{\left( 1 \right)^2} = - 8 \cr
& {\text{The derivative does not change sign, so at }}x = 0{\text{ there is}} \cr
& {\text{not an extreme}}{\text{.}} \cr
& *f''\left( 3 \right) = 12{\left( 3 \right)^2} - 24\left( 3 \right) = 36 > 0 \cr
& {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr
& f\left( x \right){\text{ has a relative minimum at }}\left( {3,f\left( 3 \right)} \right) \cr
& f\left( 3 \right) = {\left( 3 \right)^4} - 4{\left( 3 \right)^3} + 2 = - 25 \cr
& {\text{Relative minimum at }}\left( {3, - 25} \right) \cr} $$
