Cubic Graph A1F548

1. **State the problem:** We need to sketch the graph of the function $$f(x) = 2 + 3x - x^3$$. 2. **Understand the function:** This is a cubic polynomial function of the form $$f(x) = a x^3 + b x^2 + c x + d$$ where here $$a = -1$$, $$b = 0$$, $$c = 3$$, and $$d = 2$$. 3. **Find critical points:** To find where the function increases or decreases, compute the derivative: $$f'(x) = \frac{d}{dx}(2 + 3x - x^3) = 3 - 3x^2$$ Set the derivative equal to zero to find critical points: $$3 - 3x^2 = 0$$ $$\Rightarrow 3x^2 = 3$$ $$\Rightarrow x^2 = 1$$ $$\Rightarrow x = \pm 1$$ 4. **Evaluate the function at critical points:** $$f(1) = 2 + 3(1) - 1^3 = 2 + 3 - 1 = 4$$ $$f(-1) = 2 + 3(-1) - (-1)^3 = 2 - 3 + 1 = 0$$ 5. **Determine concavity:** Compute the second derivative: $$f''(x) = \frac{d}{dx}(3 - 3x^2) = -6x$$ Evaluate at critical points: $$f''(1) = -6(1) = -6 < 0$$ (local maximum at $$x=1$$) $$f''(-1) = -6(-1) = 6 > 0$$ (local minimum at $$x=-1$$) 6. **Find intercepts:** - **y-intercept:** At $$x=0$$, $$f(0) = 2 + 0 - 0 = 2$$ - **x-intercepts:** Solve $$2 + 3x - x^3 = 0$$. This cubic does not factor nicely, so approximate or use numerical methods. Roughly, the function crosses the x-axis near $$x = -1.7$$ and $$x = 1.8$$. 7. **Sketch the graph:** - The graph passes through (0,2). - It has a local minimum at (-1,0). - It has a local maximum at (1,4). - The end behavior is $$f(x) \to -\infty$$ as $$x \to \infty$$ and $$f(x) \to \infty$$ as $$x \to -\infty$$ because the leading term is $$-x^3$$. This gives a typical cubic shape flipped upside down. **Final answer:** The graph of $$f(x) = 2 + 3x - x^3$$ has local max at (1,4), local min at (-1,0), y-intercept at (0,2), and crosses the x-axis near $$x \approx -1.7$$ and $$x \approx 1.8$$.