Answer
$$\eqalign{
& {\text{Increasing on: }}\left( { - \infty ,1} \right){\text{ and }}\left( {\frac{7}{3},\infty } \right) \cr
& {\text{Decreasing on: }}\left( {1,\frac{7}{3}} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {x - 1} \right)^2}\left( {x - 3} \right) \cr
& {\text{Calculate the first derivative}} \cr
& f'\left( x \right) = \underbrace {\frac{d}{{dx}}\left[ {{{\left( {x - 1} \right)}^2}\left( {x - 3} \right)} \right]}_{{\text{Use product rule}}} \cr
& f'\left( x \right) = {\left( {x - 1} \right)^2}\left( 1 \right) + \left( {x - 3} \right)\left( 2 \right)\left( {x - 1} \right) \cr
& f'\left( x \right) = \left( {x - 1} \right)\left( {x - 1 + 2\left( {x - 3} \right)} \right) \cr
& f'\left( x \right) = \left( {x - 1} \right)\left( {3x - 7} \right) \cr
& {\text{Find the critical points, set the first derivative to }}0 \cr
& \left( {x - 1} \right)\left( {3x - 7} \right) = 0 \cr
& x = 1,{\text{ }}x = \frac{7}{3} \cr
& {\text{Set the intervals }}\left( { - \infty ,1} \right){\text{, }}\left( {1,\frac{7}{3}} \right){\text{ and }}\left( {\frac{7}{3},\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 178 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \infty ,1} \right)}&{\left( {1,\frac{7}{3}} \right)}&{\left( {\frac{7}{3},\infty } \right)} \\
{{\text{Test Value}}}&{x = 0}&{x = 2}&{x = 3} \\
{{\text{Sign of }}f'\left( x \right)}&{7 > 0}&{ - 1 < 0}&{4 > 0} \\
{{\text{Conclusion}}}&{{\text{Increasing}}}&{{\text{Decreasing}}}&{{\text{Increasing}}}
\end{array}}\]
$$\eqalign{
& {\text{By Theorem 3}}{\text{.5 }}f{\text{ is:}} \cr
& {\text{Increasing on: }}\left( { - \infty ,1} \right){\text{ and }}\left( {\frac{7}{3},\infty } \right) \cr
& {\text{Decreasing on: }}\left( {1,\frac{7}{3}} \right) \cr} $$
