Find the local maximum and minimum of $LaTeX: \displaystyle f(x) = - 9 x^{3} + 6 x^{2} - x - 4$ .

To find the critical numbers solve $LaTeX: \displaystyle f'(x) = 0$ . The derivative is $LaTeX: \displaystyle f'(x) = - 27 x^{2} + 12 x - 1$ . Solving $LaTeX: \displaystyle - 27 x^{2} + 12 x - 1 = 0$ gives $LaTeX: \displaystyle x = \left[ \frac{1}{9}, \ \frac{1}{3}\right]$ . Using the 2nd derivative test gives:
$LaTeX: \displaystyle f''\left( \frac{1}{9} \right) = 6$ which is greater than zero, so the function is concave up and $LaTeX: \displaystyle f\left(\frac{1}{9}\right) = - \frac{328}{81}$ is a local minimum.
$LaTeX: \displaystyle f''\left( \frac{1}{3} \right) = -6$ which is less than zero, so the function is concave down and $LaTeX: \displaystyle f\left(\frac{1}{3}\right) = -4$ is a local maximum.