Answer
$$\eqalign{
& {\text{Concave up on }}\left( {1,\infty } \right) \cr
& {\text{Concave down on }}\left( { - \infty ,1} \right) \cr
& {\text{The inflection points is at: }}x = 1 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {e^x}\left( {x - 3} \right) \cr
& \cr
& {\text{Calculate the second derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}\left( {x - 3} \right)} \right] \cr
& f'\left( x \right) = {e^x}\left( 1 \right) + \left( {x - 3} \right){e^x} \cr
& f'\left( x \right) = {e^x}\left( {1 + x - 3} \right) \cr
& f'\left( x \right) = {e^x}\left( {x - 2} \right) \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}\left( {x - 2} \right)} \right] \cr
& f''\left( x \right) = {e^x}\left( 1 \right) + {e^x}\left( {x - 2} \right) \cr
& f''\left( x \right) = {e^x}\left( {1 + x - 2} \right) \cr
& f''\left( x \right) = {e^x}\left( {x - 1} \right) \cr
& \cr
& {\text{Set the second derivative to 0}} \cr
& f''\left( x \right) = {e^x}\left( {x - 1} \right) = 0 \cr
& {\text{We see that }}\,f''\left( x \right) = 0{\text{ at }}x = 1 \cr
& {\text{That points is candidate for inflection points}} \cr
& {\text{We need evaluate the intervals }}\left( { - \infty ,1} \right){\text{ and }}\left( {1,\infty } \right) \cr
& \cr
& {\text{Now}}{\text{, we will evaluate test values and resume in a table}} \cr
& {\text{to determinate whether the concavity changes at these points}} \cr} $$
\[\begin{array}{*{20}{c}}
{{\text{Interval}}}&{{\text{Test value }}\left( x \right)}&{{\text{Sign of }}f''\left( x \right)}&{{\text{Behavior of }}f\left( x \right)} \\
{\left( { - \infty ,1} \right)}&0&{f''\left( 0 \right) < 0}&{{\text{Concave down}}} \\
{\left( {1,\infty } \right)}&2&{f''\left( 2 \right) > 0}&{{\text{Concave up}}}
\end{array}\]
$$\eqalign{
& {\text{From the table we can conclude that the function is:}} \cr
& {\text{Concave up on }}\left( {1,\infty } \right) \cr
& {\text{Concave down on }}\left( { - \infty ,1} \right) \cr
& {\text{The inflection points is at: }}x = 1 \cr} $$
