Answer
$$\eqalign{
& {\text{Inflection point }}\left( {2,8} \right) \cr
& {\text{Concave upward}}:{\text{ }}\left( {2,\infty } \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( { - \infty ,2} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3} - 6{x^2} + 12x \cr
& {\text{Calculate the second derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3} - 6{x^2} + 12x} \right] \cr
& f'\left( x \right) = 3{x^2} - 12x + 12 \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {3{x^2} - 12x + 12} \right] \cr
& f''\left( x \right) = 6x - 12 \cr
& {\text{Set }}f''\left( x \right) = 0 \cr
& 6x - 12 = 0 \cr
& x = 2 \cr
& {\text{Set the intervals }}\left( { - \infty ,2} \right),\left( {2,\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \infty ,2} \right)}&{\left( {2,\infty } \right)} \\
{{\text{Test Value}}}&{x = - 1}&{x = 5} \\
{{\text{Sign of }}f''\left( x \right)}&{f''\left( { - 1} \right) = - 18 < 0}&{f''\left( 5 \right) = 18 > 0} \\
{{\text{Conclusion}}}&{{\text{Concave downward}}}&{{\text{Concave upward}}}
\end{array}}\]
$$\eqalign{
& {\text{Inflection point }}\left( {2,8} \right) \cr
& {\text{Concave upward}}:{\text{ }}\left( {2,\infty } \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( { - \infty ,2} \right) \cr} $$
