1. **State the problem:** Given the function $g(x) = x^3 - 3x^2 - 9x - 5$, we are asked to find the first and second derivatives, verify stationary points, determine their nature, find the inflection point, and sketch the graph with key features. 2. **Find $g'(x)$ and $g''(x)$:** - Differentiate $g(x)$ term-by-term: $$g'(x) = 3x^2 - 6x - 9$$ - Differentiate $g'(x)$ to get $g''(x)$: $$g''(x) = 6x - 6$$ 3. **Show that the stationary points are $(-1,0)$ and $(3,-32)$:** - Stationary points occur where $g'(x) = 0$: $$3x^2 - 6x - 9 = 0$$ - Divide entire equation by 3: $$x^2 - 2x - 3 = 0$$ - Factor: $$(x - 3)(x + 1) = 0$$ - So stationary points at $x = 3$ and $x = -1$. - Evaluate $g(-1)$: $$(-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3 + 9 - 5 = 0$$ - Evaluate $g(3)$: $$3^3 - 3(3)^2 - 9(3) - 5 = 27 - 27 - 27 - 5 = -32$$ 4. **Determine the nature of each stationary point:** - Use $g''(x)$: - At $x = -1$: $$g''(-1) = 6(-1) - 6 = -6 - 6 = -12 < 0$$ So $(-1,0)$ is a local maximum. - At $x = 3$: $$g''(3) = 6(3) - 6 = 18 - 6 = 12 > 0$$ So $(3,-32)$ is a local minimum. 5. **Find the inflection point:** - Inflection point occurs where $g''(x) = 0$: $$6x - 6 = 0 \,\Rightarrow\, x = 1$$ - Evaluate $g(1)$: $$1^3 - 3(1)^2 - 9(1) - 5 = 1 - 3 - 9 - 5 = -16$$ - Inflection point is at $(1, -16)$. 6. **Graph features summary:** - Stationary points: $(-1, 0)$ local max and $(3, -32)$ local min. - Inflection point: $(1, -16)$. - X-intercepts at $(-1, 0)$ and $(5, 0)$. Final answer: - $$g'(x) = 3x^2 - 6x - 9$$ - $$g''(x) = 6x - 6$$ - Stationary points: $(-1,0)$ (max), $(3,-32)$ (min) - Inflection point: $(1,-16)$