Find the absolute maximum of $LaTeX: \displaystyle f(x) = - \frac{20 x^{3}}{343} - \frac{90 x^{2}}{343} + \frac{600 x}{343} + \frac{692}{343}$ on $LaTeX: \displaystyle [-7,4]$

Taking the derivative gives $LaTeX: \displaystyle f'(x) = - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343}$ . Setting it equal to zero and solving gives the critical numbers. $LaTeX: \displaystyle - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343} = 0$ . The critical numbers are $LaTeX: \displaystyle x = -5$ 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 $LaTeX: \displaystyle {-7, 2, -5, 4}$ and evaluating gives $LaTeX: \displaystyle \left( -7, \ - \frac{1058}{343}\right), \left( 2, \ 4\right), \left( -5, \ -6\right), \left( 4, \ \frac{372}{343}\right)$ . The max is $LaTeX: \displaystyle \left( 2, \ 4\right)$ and the min is $LaTeX: \displaystyle \left( -5, \ -6\right)$ .