Find the absolute maximum of $LaTeX: \displaystyle f(x) = \frac{6 x^{3}}{125} + \frac{9 x^{2}}{125} - \frac{108 x}{125} - \frac{1118}{125}$ on $LaTeX: \displaystyle [-9,9]$

Taking the derivative gives $LaTeX: \displaystyle f'(x) = \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125}$ . Setting it equal to zero and solving gives the critical numbers. $LaTeX: \displaystyle \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125} = 0$ . The critical numbers are $LaTeX: \displaystyle x = -3$ and $LaTeX: \displaystyle x = 2$ . The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are $LaTeX: \displaystyle {9, 2, -3, -9}$ and evaluating gives $LaTeX: \displaystyle \left( 9, \ \frac{3013}{125}\right), \left( 2, \ -10\right), \left( -3, \ -7\right), \left( -9, \ - \frac{3791}{125}\right)$ . The max is $LaTeX: \displaystyle \left( 9, \ \frac{3013}{125}\right)$ and the min is $LaTeX: \displaystyle \left( -9, \ - \frac{3791}{125}\right)$ .