1. **Problem statement:** Given the function $g(x) = \frac{1}{9}x^3 - 3x$, analyze its properties as requested in parts a) to g). 2. **Part a) Graph sketch:** The function is cubic with leading term $\frac{1}{9}x^3$ and a linear term $-3x$. It has inflection and intercepts near $x = -3\sqrt{3}, 0, 3\sqrt{3}$. Use a CAS app to plot and sketch accordingly. 3. **Part b) Monotonicity intervals:** Calculate the derivative: $$g'(x) = \frac{d}{dx}\left(\frac{1}{9}x^3 - 3x\right) = \frac{1}{3}x^2 - 3$$ Set $g'(x) = 0$ to find critical points: $$\frac{1}{3}x^2 - 3 = 0 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$$ Test intervals: - For $x < -3$, choose $x = -4$: $g'(-4) = \frac{1}{3}16 - 3 = \frac{16}{3} - 3 > 0$ increasing - For $-3 < x < 3$, choose $x=0$: $g'(0) = -3 < 0$ decreasing - For $x > 3$, choose $x=4$: $g'(4) = \frac{16}{3} - 3 > 0$ increasing **Answer:** - Strictly increasing on $(-\infty, -3)$ and $(3, \infty)$ - Strictly decreasing on $(-3, 3)$ 4. **Part c) Zeros and y-intercept:** Set $g(x) = 0$: $$\frac{1}{9}x^3 - 3x = 0 \Rightarrow x\left(\frac{1}{9}x^2 - 3\right) = 0$$ So zeros are: $$x=0 \quad \text{or} \quad \frac{1}{9}x^2 - 3 = 0 \Rightarrow x^2 = 27 \Rightarrow x = \pm 3\sqrt{3}$$ Y-intercept is $g(0) = 0$. 5. **Part d) Check if point $A(3, -6)$ lies on graph:** Calculate $g(3)$: $$g(3) = \frac{1}{9} \cdot 27 - 3 \cdot 3 = 3 - 9 = -6$$ Since $g(3) = -6$, point $A$ lies on the graph. 6. **Part e) Derivative and evaluation at 3:** Already found: $$g'(x) = \frac{1}{3}x^2 - 3$$ Calculate $g'(3)$: $$g'(3) = \frac{1}{3} \cdot 9 - 3 = 3 - 3 = 0$$ 7. **Part f) Tangent equation at $A(3, -6)$:** Tangent line formula: $$y = g'(3)(x - 3) + g(3)$$ Since $g'(3) = 0$ and $g(3) = -6$: $$y = 0 \cdot (x - 3) - 6 = -6$$ So tangent is the horizontal line: $$y = -6$$ 8. **Part g) Extremes and their nature:** Critical points at $x = \pm 3$ from $g'(x) = 0$. Second derivative: $$g''(x) = \frac{d}{dx}g'(x) = \frac{d}{dx}\left(\frac{1}{3}x^2 - 3\right) = \frac{2}{3}x$$ Evaluate at critical points: - At $x = -3$: $$g''(-3) = \frac{2}{3} \cdot (-3) = -2 < 0$$ So $x = -3$ is a local maximum. - At $x = 3$: $$g''(3) = \frac{2}{3} \cdot 3 = 2 > 0$$ So $x = 3$ is a local minimum. Calculate coordinates: - $g(-3) = \frac{1}{9}(-3)^3 - 3(-3) = \frac{1}{9}(-27) + 9 = -3 + 9 = 6$ - $g(3) = -6$ (already calculated) **Extremes:** - Local max at $(-3, 6)$ - Local min at $(3, -6)$