1. **State the problem:** Find the extreme values of the function $f(x) = x^3 - 18x^2 + 96$ using the second derivative test. 2. **Find the first derivative:** $$f'(x) = \frac{d}{dx}(x^3 - 18x^2 + 96) = 3x^2 - 36x$$ 3. **Find critical points by setting $f'(x) = 0$:** $$3x^2 - 36x = 0$$ $$3x(x - 12) = 0$$ Critical points are $x = 0$ and $x = 12$. 4. **Find the second derivative:** $$f''(x) = \frac{d}{dx}(3x^2 - 36x) = 6x - 36$$ 5. **Apply the second derivative test:** - At $x=0$: $$f''(0) = 6(0) - 36 = -36 < 0$$ Since $f''(0) < 0$, $f$ has a local maximum at $x=0$. - At $x=12$: $$f''(12) = 6(12) - 36 = 72 - 36 = 36 > 0$$ Since $f''(12) > 0$, $f$ has a local minimum at $x=12$. 6. **Find the extreme values by evaluating $f(x)$ at critical points:** - $f(0) = 0^3 - 18(0)^2 + 96 = 96$ - $f(12) = 12^3 - 18(12)^2 + 96 = 1728 - 2592 + 96 = -768$ **Final answer:** - Local maximum value is $96$ at $x=0$. - Local minimum value is $-768$ at $x=12$.