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 of $f$ is concave up or concave down and the inflection points. 2. **Step i) Increasing/Decreasing and Local Extrema:** - The first derivative $f'(x)$ gives the slope of the function: $$f'(x) = \frac{d}{dx}\left(-\frac{1}{3}x^3 + 2x^2 - 12\right) = -x^2 + 4x$$ - To find critical points, solve $f'(x) = 0$: $$-x^2 + 4x = 0$$ $$x(-x + 4) = 0$$ $$x = 0 \quad \text{or} \quad x = 4$$ - Test intervals around critical points to determine increasing/decreasing: For $x < 0$, pick $x = -1$: $$f'(-1) = -(-1)^2 + 4(-1) = -1 - 4 = -5 < 0$$ (decreasing) For $0 < x < 4$, pick $x = 2$: $$f'(2) = -(2)^2 + 4(2) = -4 + 8 = 4 > 0$$ (increasing) For $x > 4$, pick $x = 5$: $$f'(5) = -(5)^2 + 4(5) = -25 + 20 = -5 < 0$$ (decreasing) - Therefore: - $f$ is decreasing on $(-\infty, 0)$ - $f$ is increasing on $(0, 4)$ - $f$ is decreasing on $(4, \infty)$ - Local extrema occur at critical points where the derivative changes sign: - At $x=0$, $f'$ changes from negative to positive, so local minimum. - At $x=4$, $f'$ changes from positive to negative, so local maximum. - Find the function values at these points: $$f(0) = -\frac{1}{3}(0)^3 + 2(0)^2 - 12 = -12$$ $$f(4) = -\frac{1}{3}(4)^3 + 2(4)^2 - 12 = -\frac{1}{3}(64) + 2(16) - 12 = -\frac{64}{3} + 32 - 12 = -\frac{64}{3} + 20 = \frac{-64 + 60}{3} = -\frac{4}{3}$$ 3. **Step ii) Concavity and Inflection Points:** - The second derivative $f''(x)$ gives concavity: $$f''(x) = \frac{d}{dx}f'(x) = \frac{d}{dx}(-x^2 + 4x) = -2x + 4$$ - Find inflection points by solving $f''(x) = 0$: $$-2x + 4 = 0$$ $$-2x = -4$$ $$x = 2$$ - Test concavity intervals: For $x < 2$, pick $x=1$: $$f''(1) = -2(1) + 4 = 2 > 0$$ (concave up) For $x > 2$, pick $x=3$: $$f''(3) = -2(3) + 4 = -6 + 4 = -2 < 0$$ (concave down) - Inflection point at $x=2$ where concavity changes. - Find $f(2)$: $$f(2) = -\frac{1}{3}(2)^3 + 2(2)^2 - 12 = -\frac{8}{3} + 8 - 12 = -\frac{8}{3} - 4 = -\frac{8}{3} - \frac{12}{3} = -\frac{20}{3}$$ **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 $(2, -\frac{20}{3})$.