Find the local maximum and minimum of $LaTeX: \displaystyle f(x) = x^{3} + 6 x^{2} + 9 x + 1$ .

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