1. **Problem statement:** Given the function $f(x) = -\frac{1}{3}x^3 + 2x^2 - 12$, find: i) The intervals where $f$ is increasing or decreasing and the local maxima and minima. ii) The intervals where the graph is concave up or down and the inflection points. 2. **Formulas and rules:** - To find increasing/decreasing intervals, compute the first derivative $f'(x)$ and analyze its sign. - Local maxima occur where $f'(x)$ changes from positive to negative; local minima where it changes from negative to positive. - To find concavity, compute the second derivative $f''(x)$. - The graph is concave up where $f''(x) > 0$ and concave down where $f''(x) < 0$. - Inflection points occur where $f''(x) = 0$ and the concavity changes. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx}\left(-\frac{1}{3}x^3 + 2x^2 - 12\right) = -x^2 + 4x$$ 4. **Find critical points by setting $f'(x) = 0$:** $$-x^2 + 4x = 0$$ $$x(-x + 4) = 0$$ $$x = 0 \quad \text{or} \quad x = 4$$ 5. **Determine intervals of increase/decrease by testing values around critical points:** - For $x < 0$, choose $x = -1$: $f'(-1) = -(-1)^2 + 4(-1) = -1 - 4 = -5 < 0$ (decreasing) - For $0 < x < 4$, choose $x = 2$: $f'(2) = -(2)^2 + 4(2) = -4 + 8 = 4 > 0$ (increasing) - For $x > 4$, choose $x = 5$: $f'(5) = -(5)^2 + 4(5) = -25 + 20 = -5 < 0$ (decreasing) 6. **Local maxima and minima:** - At $x=0$, $f'(x)$ changes from negative to positive, so local minimum. - At $x=4$, $f'(x)$ changes from positive to negative, so local maximum. 7. **Find the second derivative:** $$f''(x) = \frac{d}{dx}(-x^2 + 4x) = -2x + 4$$ 8. **Find possible inflection points by setting $f''(x) = 0$:** $$-2x + 4 = 0$$ $$2x = 4$$ $$x = 2$$ 9. **Determine concavity intervals by testing values around $x=2$:** - For $x < 2$, choose $x=1$: $f''(1) = -2(1) + 4 = 2 > 0$ (concave up) - For $x > 2$, choose $x=3$: $f''(3) = -2(3) + 4 = -6 + 4 = -2 < 0$ (concave down) 10. **Find function values at critical and inflection points:** - $f(0) = -\frac{1}{3}(0)^3 + 2(0)^2 - 12 = -12$ - $f(4) = -\frac{1}{3}(64) + 2(16) - 12 = -\frac{64}{3} + 32 - 12 = -\frac{64}{3} + 20 = \frac{-64 + 60}{3} = -\frac{4}{3}$ - $f(2) = -\frac{1}{3}(8) + 2(4) - 12 = -\frac{8}{3} + 8 - 12 = -\frac{8}{3} - 4 = -\frac{8 + 12}{3} = -\frac{20}{3}$ **Final answers:** - Increasing on $(0,4)$ - Decreasing on $(-\infty,0)$ and $(4,\infty)$ - Local minimum at $(0,-12)$ - Local maximum at $(4,-\frac{4}{3})$ - Concave up on $(-\infty,2)$ - Concave down on $(2,\infty)$ - Inflection point at $(2,-\frac{20}{3})$