Answer
$$\eqalign{
& \left( {\text{a}} \right)x = 3 \cr
& \left( {\text{b}} \right){\text{Decreasing: }}\left( { - \infty ,3} \right),{\text{ Increasing: }}\left( {3,\infty } \right) \cr
& \left( {\text{c}} \right){\text{Relative minimum at }}\left( {0, - 4} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^2} - 6x + 5 \cr
& \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^2} - 6x + 5} \right] \cr
& f'\left( x \right) = 2x - 6 \cr
& {\text{Calculating the critical points, set }}f'\left( x \right) = 0 \cr
& f'\left( x \right) = 0 \cr
& 2x - 6 = 0 \cr
& x = 3 \cr
& \cr
& \left( {\text{b}} \right){\text{Set the intervals }}\left( { - \infty ,3} \right),\left( {3,\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$
\[\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \infty ,3} \right)}&{\left( {3,\infty } \right)} \\
{{\text{Test Value}}}&{x = - 5}&{x = 4} \\
{{\text{Sign of }}f'\left( x \right)}&{{\text{ }}f'\left( { - 5} \right) = - 16 < 0}&{{\text{ }}f'\left( 4 \right) = 2 > 0} \\
{{\text{Conclusion}}}&{{\text{Decreasing}}}&{{\text{Increasing}}}
\end{array}\]
$$\eqalign{
& \left( {\text{c}} \right){\text{ By Theorem 3}}{\text{.6}} \cr
& f'\left( x \right){\text{ changes from negative to positive at }}x = 3,{\text{ so }}f\left( x \right) \cr
& {\text{has a relative minimum at }}\left( {3,f\left( 3 \right)} \right) \cr
& f\left( 3 \right) = {\left( 3 \right)^2} - 6\left( 3 \right) + 5 \cr
& f\left( 3 \right) = - 4 \cr
& {\text{Relative minimum at }}\left( {3, - 4} \right) \cr
& \cr
& \left( {\text{d}} \right){\text{Graph}} \cr} $$
