1. **Problem:** Given the function $$f(x) = \frac{1}{2} x^2 e^{x+1}$$ 2. **Find the domain of $$f(x)$$:** Since $$f(x)$$ is composed of polynomial $$x^2$$ and exponential $$e^{x+1}$$, both defined for all real numbers, the domain is all real numbers. **Domain:** $$\mathbb{R}$$ 3. **Find relative asymptotes (if any):** As $$x \to \pm \infty$$, $$e^{x+1}$$ grows exponentially, and $$x^2$$ grows polynomially. - As $$x \to +\infty$$, $$f(x) \to +\infty$$ (no horizontal or oblique asymptotes). - As $$x \to -\infty$$, $$e^{x+1} \to 0$$ exponentially, and $$x^2$$ grows large, but the exponential dominates, so $$f(x) \to 0$$. No vertical asymptotes since denominator is 1. **Relative asymptotes:** None. 4. **Study first and second derivatives with variation table:** - First derivative: $$f(x) = \frac{1}{2} x^2 e^{x+1}$$ Using product rule: $$f'(x) = \frac{1}{2} \left(2x e^{x+1} + x^2 e^{x+1} \right) = \frac{1}{2} e^{x+1} (2x + x^2) = \frac{1}{2} e^{x+1} x (x + 2)$$ - Second derivative: $$f''(x) = \frac{1}{2} \frac{d}{dx} \left(e^{x+1} x (x+2) \right)$$ Apply product rule again: $$f''(x) = \frac{1}{2} \left(e^{x+1} x (x+2) + e^{x+1} \frac{d}{dx} [x (x+2)] \right)$$ Calculate derivative inside: $$\frac{d}{dx} [x (x+2)] = \frac{d}{dx} (x^2 + 2x) = 2x + 2$$ So: $$f''(x) = \frac{1}{2} e^{x+1} \left(x (x+2) + 2x + 2 \right) = \frac{1}{2} e^{x+1} (x^2 + 2x + 2x + 2) = \frac{1}{2} e^{x+1} (x^2 + 4x + 2)$$ - **Critical points:** Solve $$f'(x) = 0$$ $$\frac{1}{2} e^{x+1} x (x+2) = 0 \implies x=0 \text{ or } x=-2$$ - **Sign of $$f'(x)$$:** For $$x < -2$$, $$x$$ and $$x+2$$ are negative and negative, product positive, so $$f'(x) > 0$$. For $$-2 < x < 0$$, $$x$$ negative, $$x+2$$ positive, product negative, so $$f'(x) < 0$$. For $$x > 0$$, both positive, product positive, so $$f'(x) > 0$$. - **Variation table:** | Interval | $$(-\infty, -2)$$ | $$-2$$ | $$(-2, 0)$$ | $$0$$ | $$(0, +\infty)$$ | |---|---|---|---|---|---| | $$f'(x)$$ | + | 0 | - | 0 | + | | $$f(x)$$ | Increasing | Max | Decreasing | Min | Increasing | - **Concavity:** Solve $$f''(x) = 0$$ $$x^2 + 4x + 2 = 0$$ Use quadratic formula: $$x = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm \sqrt{8}}{2} = -2 \pm \sqrt{2}$$ - For $$x < -2 - \sqrt{2}$$, $$f''(x) > 0$$ (concave up) - For $$-2 - \sqrt{2} < x < -2 + \sqrt{2}$$, $$f''(x) < 0$$ (concave down) - For $$x > -2 + \sqrt{2}$$, $$f''(x) > 0$$ (concave up) 5. **Sketch the curve:** - Domain: all real numbers - No asymptotes - Critical points at $$x=-2$$ (local max), $$x=0$$ (local min) - Inflection points at $$x = -2 \pm \sqrt{2}$$ This completes the solution for the first problem. **Final answers:** - Domain: $$\mathbb{R}$$ - No asymptotes - Critical points: $$x=-2$$ (max), $$x=0$$ (min) - Inflection points: $$x = -2 \pm \sqrt{2}$$