1. **State the problem:** Find the extreme values of the function $f(x) = x^3 - 18x^2 + 96$ using the second derivative test. 2. **Recall the formulas and rules:** - The first derivative $f'(x)$ gives critical points where $f'(x) = 0$. - The second derivative $f''(x)$ helps classify these points: - If $f''(x) > 0$, the point is a local minimum. - If $f''(x) < 0$, the point is a local maximum. - If $f''(x) = 0$, the test is inconclusive. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx}(x^3 - 18x^2 + 96) = 3x^2 - 36x$$ 4. **Find critical points by solving $f'(x) = 0$:** $$3x^2 - 36x = 0$$ $$3x(x - 12) = 0$$ So, $x = 0$ or $x = 12$ are critical points. 5. **Find the second derivative:** $$f''(x) = \frac{d}{dx}(3x^2 - 36x) = 6x - 36$$ 6. **Evaluate $f''(x)$ at critical points:** - At $x=0$: $f''(0) = 6(0) - 36 = -36 < 0$, so $x=0$ is a local maximum. - At $x=12$: $f''(12) = 6(12) - 36 = 72 - 36 = 36 > 0$, so $x=12$ is a local minimum. 7. **Find the extreme values by plugging back into $f(x)$:** - $f(0) = 0^3 - 18(0)^2 + 96 = 96$ - $f(12) = 12^3 - 18(12)^2 + 96 = 1728 - 2592 + 96 = -768$ **Final answer:** - Local maximum at $x=0$ with value $96$. - Local minimum at $x=12$ with value $-768$.