Find the absolute maximum of $LaTeX: \displaystyle f(x) = \frac{8 x^{3}}{9} - \frac{4 x^{2}}{3} - \frac{16 x}{3} + \frac{62}{9}$ on $LaTeX: \displaystyle [-4,4]$

Taking the derivative gives $LaTeX: \displaystyle f'(x) = \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3}$ . Setting it equal to zero and solving gives the critical numbers. $LaTeX: \displaystyle \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3} = 0$ . The critical numbers are $LaTeX: \displaystyle x = -1$ 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 $LaTeX: \displaystyle {2, -4, 4, -1}$ and evaluating gives $LaTeX: \displaystyle \left( 2, \ -2\right), \left( -4, \ -50\right), \left( 4, \ \frac{190}{9}\right), \left( -1, \ 10\right)$ . The max is $LaTeX: \displaystyle \left( 4, \ \frac{190}{9}\right)$ and the min is $LaTeX: \displaystyle \left( -4, \ -50\right)$ .