1. **State the problem:** We have the function $f(x) = -\frac{1}{3}x^3 + 2x^2 - 12$. We need to find: i) Intervals where $f$ is increasing or decreasing and local maxima/minima. ii) Intervals where $f$ is concave up or down and inflection points. 2. **Find the first derivative $f'(x)$ to analyze increasing/decreasing behavior:** $$f'(x) = \frac{d}{dx}\left(-\frac{1}{3}x^3 + 2x^2 - 12\right) = -x^2 + 4x$$ 3. **Find critical points by setting $f'(x) = 0$:** $$-x^2 + 4x = 0$$ $$x(-x + 4) = 0$$ $$x = 0 \quad \text{or} \quad x = 4$$ 4. **Determine intervals of increase/decrease by testing $f'(x)$ sign:** - For $x < 0$, choose $x = -1$: $f'(-1) = -1 + (-4) = -5 < 0$ (decreasing) - For $0 < x < 4$, choose $x = 2$: $f'(2) = -4 + 8 = 4 > 0$ (increasing) - For $x > 4$, choose $x = 5$: $f'(5) = -25 + 20 = -5 < 0$ (decreasing) 5. **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. 6. **Find $f(0)$ and $f(4)$ for local extrema values:** $$f(0) = -\frac{1}{3}(0)^3 + 2(0)^2 - 12 = -12$$ $$f(4) = -\frac{1}{3}(64) + 2(16) - 12 = -\frac{64}{3} + 32 - 12 = -\frac{64}{3} + 20 = \frac{-64 + 60}{3} = -\frac{4}{3}$$ 7. **Find the second derivative $f''(x)$ to analyze concavity:** $$f''(x) = \frac{d}{dx}(-x^2 + 4x) = -2x + 4$$ 8. **Find inflection points by setting $f''(x) = 0$:** $$-2x + 4 = 0$$ $$2x = 4$$ $$x = 2$$ 9. **Determine concavity intervals by testing $f''(x)$ sign:** - For $x < 2$, choose $x=0$: $f''(0) = 4 > 0$ (concave up) - For $x > 2$, choose $x=3$: $f''(3) = -6 + 4 = -2 < 0$ (concave down) 10. **Find $f(2)$ for inflection point coordinates:** $$f(2) = -\frac{1}{3}(8) + 2(4) - 12 = -\frac{8}{3} + 8 - 12 = -\frac{8}{3} - 4 = -\frac{8 + 12}{3} = -\frac{20}{3}$$ **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 $(2,-\frac{20}{3})$.