Find the absolute maximum of $LaTeX: \displaystyle f(x) = - \frac{8 x^{3}}{125} + \frac{36 x^{2}}{125} + \frac{96 x}{125} - \frac{448}{125}$ on $LaTeX: \displaystyle [-2,10]$

Taking the derivative gives $LaTeX: \displaystyle f'(x) = - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125}$ . Setting it equal to zero and solving gives the critical numbers. $LaTeX: \displaystyle - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125} = 0$ . The critical numbers are $LaTeX: \displaystyle x = -1$ and $LaTeX: \displaystyle x = 4$ . The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are $LaTeX: \displaystyle {10, 4, -2, -1}$ and evaluating gives $LaTeX: \displaystyle \left( 10, \ - \frac{3888}{125}\right), \left( 4, \ 0\right), \left( -2, \ - \frac{432}{125}\right), \left( -1, \ -4\right)$ . The max is $LaTeX: \displaystyle \left( 4, \ 0\right)$ and the min is $LaTeX: \displaystyle \left( 10, \ - \frac{3888}{125}\right)$ .