Find the absolute maximum of $LaTeX: \displaystyle f(x) = \frac{7 x^{3}}{256} + \frac{21 x^{2}}{256} - \frac{315 x}{256} - \frac{1993}{256}$ on $LaTeX: \displaystyle [-7,5]$

Taking the derivative gives $LaTeX: \displaystyle f'(x) = \frac{21 x^{2}}{256} + \frac{21 x}{128} - \frac{315}{256}$ . Setting it equal to zero and solving gives the critical numbers. $LaTeX: \displaystyle \frac{21 x^{2}}{256} + \frac{21 x}{128} - \frac{315}{256} = 0$ . The critical numbers are $LaTeX: \displaystyle x = -5$ 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 $LaTeX: \displaystyle {3, -7, -5, 5}$ and evaluating gives $LaTeX: \displaystyle \left( 3, \ -10\right), \left( -7, \ - \frac{145}{32}\right), \left( -5, \ -3\right), \left( 5, \ - \frac{271}{32}\right)$ . The max is $LaTeX: \displaystyle \left( -5, \ -3\right)$ and the min is $LaTeX: \displaystyle \left( 3, \ -10\right)$ .