Find the absolute maximum of $LaTeX: \displaystyle f(x) = - \frac{11 x^{3}}{32} + \frac{33 x^{2}}{32} + \frac{99 x}{32} - \frac{9}{32}$ on $LaTeX: \displaystyle [-9,5]$

Taking the derivative gives $LaTeX: \displaystyle f'(x) = - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32}$ . Setting it equal to zero and solving gives the critical numbers. $LaTeX: \displaystyle - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32} = 0$ . The critical numbers are $LaTeX: \displaystyle x = -1$ and $LaTeX: \displaystyle x = 3$ . The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are $LaTeX: \displaystyle {3, 5, -1, -9}$ and evaluating gives $LaTeX: \displaystyle \left( 3, \ 9\right), \left( 5, \ -2\right), \left( -1, \ -2\right), \left( -9, \ 306\right)$ . The max is $LaTeX: \displaystyle \left( -9, \ 306\right)$ and the graph has a minimum value of $LaTeX: \displaystyle -2$ at the points $LaTeX: \displaystyle x = -1,5$