Find the absolute maximum of $LaTeX: \displaystyle f(x) = \frac{11 x^{3}}{32} - \frac{33 x^{2}}{32} - \frac{99 x}{32} + \frac{265}{32}$ on $LaTeX: \displaystyle [-4,10]$

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 $LaTeX: \displaystyle {10, 3, -4, -1}$ and evaluating gives $LaTeX: \displaystyle \left( 10, \ \frac{6975}{32}\right), \left( 3, \ -1\right), \left( -4, \ - \frac{571}{32}\right), \left( -1, \ 10\right)$ . The max is $LaTeX: \displaystyle \left( 10, \ \frac{6975}{32}\right)$ and the min is $LaTeX: \displaystyle \left( -4, \ - \frac{571}{32}\right)$ .