Answer
$$\eqalign{
& {\text{Increasing on: }}\left( { - \infty , - \frac{3}{2}} \right) \cr
& {\text{Decreasing on: }}\left( { - \frac{3}{2},\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^2} + 3x - 12 \cr
& {\text{Calculate the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^2} + 3x - 12} \right] \cr
& f'\left( x \right) = 2x + 3 \cr
& {\text{Find the critical points, set the first derivative to }}0 \cr
& f'\left( x \right) = 0 \cr
& 2x + 3 = 0 \cr
& 2x = - 3 \cr
& x = - \frac{3}{2} \cr
& {\text{Set the intervals }}\left( { - \infty , - \frac{3}{2}} \right){\text{ and }}\left( { - \frac{3}{2},\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 178 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \infty , - \frac{3}{2}} \right)}&{\left( { - \frac{3}{2},\infty } \right)} \\
{{\text{Test Value}}}&{x = - 5}&{x = 0} \\
{{\text{Sign of }}f'\left( x \right)}&{f'\left( { - 5} \right) = - 7 < 0}&{f'\left( 0 \right) = 3 > 0} \\
{{\text{Conclusion}}}&{{\text{Decreasing}}}&{{\text{Increasing}}}
\end{array}}\]
$$\eqalign{
& {\text{By Theorem 3}}{\text{.5 }}f{\text{ is:}} \cr
& {\text{Increasing on: }}\left( { - \infty , - \frac{3}{2}} \right) \cr
& {\text{Decreasing on: }}\left( { - \frac{3}{2},\infty } \right) \cr} $$
