Find the absolute maximum of $LaTeX: \displaystyle f(x) = - \frac{2 x^{3}}{49} + \frac{15 x^{2}}{49} + \frac{36 x}{49} - \frac{422}{49}$ on $LaTeX: \displaystyle [-3,7]$

Taking the derivative gives $LaTeX: \displaystyle f'(x) = - \frac{6 x^{2}}{49} + \frac{30 x}{49} + \frac{36}{49}$ . Setting it equal to zero and solving gives the critical numbers. $LaTeX: \displaystyle - \frac{6 x^{2}}{49} + \frac{30 x}{49} + \frac{36}{49} = 0$ . The critical numbers are $LaTeX: \displaystyle x = -1$ and $LaTeX: \displaystyle x = 6$ . The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are $LaTeX: \displaystyle {6, -3, -1, 7}$ and evaluating gives $LaTeX: \displaystyle \left( 6, \ -2\right), \left( -3, \ - \frac{341}{49}\right), \left( -1, \ -9\right), \left( 7, \ - \frac{121}{49}\right)$ . The max is $LaTeX: \displaystyle \left( 6, \ -2\right)$ and the min is $LaTeX: \displaystyle \left( -1, \ -9\right)$ .