Find the absolute maximum of $LaTeX: \displaystyle f(x) = \frac{x^{3}}{16} + \frac{3 x^{2}}{8} - \frac{9 x}{4} - \frac{7}{2}$ on $LaTeX: \displaystyle [-6,6]$

Taking the derivative gives $LaTeX: \displaystyle f'(x) = \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4}$ . Setting it equal to zero and solving gives the critical numbers. $LaTeX: \displaystyle \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4} = 0$ . The critical numbers are $LaTeX: \displaystyle x = -6$ and $LaTeX: \displaystyle x = 2$ . The absolute maximum is either at a critical number or at the end point of the interval. The critical number $LaTeX: \displaystyle x = -6$ is not in the interval $LaTeX: \displaystyle (-6,6)$ and does not need to be checked. The inputs to be checked are $LaTeX: \displaystyle {-6, 2, 6}$ and evaluating gives $LaTeX: \displaystyle \left( -6, \ 10\right), \left( 2, \ -6\right), \left( 6, \ 10\right)$ . The graph has a maximum value of $LaTeX: \displaystyle 10$ at the points $LaTeX: \displaystyle x = 6,-6$ and the min is $LaTeX: \displaystyle \left( 2, \ -6\right)$ .