Find the absolute maximum of $LaTeX: \displaystyle f(x) = \frac{2 x^{3}}{49} - \frac{15 x^{2}}{49} - \frac{36 x}{49} - \frac{166}{49}$ on $LaTeX: \displaystyle [-7,9]$

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 $LaTeX: \displaystyle {-7, 6, 9, -1}$ and evaluating gives $LaTeX: \displaystyle \left( -7, \ - \frac{1335}{49}\right), \left( 6, \ -10\right), \left( 9, \ - \frac{247}{49}\right), \left( -1, \ -3\right)$ . The max is $LaTeX: \displaystyle \left( -1, \ -3\right)$ and the min is $LaTeX: \displaystyle \left( -7, \ - \frac{1335}{49}\right)$ .