1. **Problem:** Given the function $$f(x) = \frac{1}{2} x^2 e^{x+1}$$ **a. Find the domain of $$f(x)$$:** 1. The function involves $$x^2$$ and $$e^{x+1}$$. 2. The exponential function $$e^{x+1}$$ is defined for all real $$x$$. 3. The polynomial $$x^2$$ is defined for all real $$x$$. 4. Therefore, the domain of $$f(x)$$ is all real numbers: $$\mathbb{R}$$. **b. Find relative asymptotes (if any):** 1. Asymptotes occur where the function tends to infinity or a line as $$x \to \pm \infty$$ or near points of discontinuity. 2. Since $$f(x)$$ is continuous and defined everywhere, no vertical asymptotes. 3. Check limits at infinity: $$\lim_{x \to +\infty} \frac{1}{2} x^2 e^{x+1} = +\infty$$ (exponential dominates) $$\lim_{x \to -\infty} \frac{1}{2} x^2 e^{x+1} = 0$$ (exponential tends to 0 faster than polynomial grows) 4. No horizontal or oblique asymptotes since function grows exponentially. 5. **Conclusion:** No relative asymptotes. **c. Study the first and second derivatives with variation table:** 1. First derivative: $$f(x) = \frac{1}{2} x^2 e^{x+1}$$ Use product rule: $$f'(x) = \frac{1}{2} \left(2x e^{x+1} + x^2 e^{x+1} \right) = \frac{1}{2} e^{x+1} (2x + x^2) = \frac{1}{2} e^{x+1} x (x + 2)$$ 2. Find critical points by setting $$f'(x) = 0$$: $$\frac{1}{2} e^{x+1} x (x + 2) = 0$$ Since $$e^{x+1} \neq 0$$, solve: $$x = 0$$ or $$x = -2$$ 3. Second derivative: $$f'(x) = \frac{1}{2} e^{x+1} x (x + 2)$$ Use product rule again: $$f''(x) = \frac{1}{2} e^{x+1} x (x + 2) + \frac{1}{2} e^{x+1} (2x + 2) = \frac{1}{2} e^{x+1} \left[x(x+2) + 2(x+1)\right] = \frac{1}{2} e^{x+1} (x^2 + 2x + 2x + 2) = \frac{1}{2} e^{x+1} (x^2 + 4x + 2)$$ 4. Find inflection points by solving $$f''(x) = 0$$: $$\frac{1}{2} e^{x+1} (x^2 + 4x + 2) = 0$$ Since $$e^{x+1} \neq 0$$, solve: $$x^2 + 4x + 2 = 0$$ Use quadratic formula: $$x = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm \sqrt{8}}{2} = -2 \pm \sqrt{2}$$ 5. Variation table summary: | Interval | Sign of $$f'(x)$$ | Increasing/Decreasing | Sign of $$f''(x)$$ | Concavity | |----------|------------------|---------------------|------------------|-----------| | $$(-\infty, -2)$$ | $$f'(-3) = \frac{1}{2} e^{-2} (-3)(-1) > 0$$ | Increasing | $$f''(-3) > 0$$ | Concave up | | $$(-2, 0)$$ | $$f'(-1) = \frac{1}{2} e^{0} (-1)(1) < 0$$ | Decreasing | $$f''(-1) < 0$$ | Concave down | | $$(0, +\infty)$$ | $$f'(1) = \frac{1}{2} e^{2} (1)(3) > 0$$ | Increasing | $$f''(1) > 0$$ | Concave up | **d. Sketch the curve:** - The function is defined everywhere. - It has critical points at $$x = -2$$ (local max) and $$x = 0$$ (local min). - The function tends to 0 as $$x \to -\infty$$ and grows exponentially as $$x \to +\infty$$. - Concavity changes at $$x = -2 \pm \sqrt{2}$$. --- **Summary:** - Domain: $$\mathbb{R}$$ - No asymptotes - Critical points at $$x = -2, 0$$ - Inflection points at $$x = -2 \pm \sqrt{2}$$ --- **Note:** The second problem is ignored as per instructions.