1. **Stating the problem:** We analyze the function $$y = xe^{-x^2}$$ to find its domain, intercepts with axes, positivity intervals, limits, and study the first and second derivatives in detail. 2. **Domain:** The function is defined for all real numbers since the exponential function is defined everywhere and multiplication by $x$ is also defined everywhere. $$\text{Domain} = (-\infty, +\infty)$$ 3. **Intercepts:** - **y-intercept:** Set $x=0$: $$y(0) = 0 \cdot e^{0} = 0$$ So the graph passes through the origin $(0,0)$. - **x-intercepts:** Set $y=0$: $$xe^{-x^2} = 0 \implies x=0$$ Since $e^{-x^2} \neq 0$ for any $x$, the only root is at $x=0$. 4. **Positivity:** Since $e^{-x^2} > 0$ for all $x$, the sign of $y$ depends on $x$: - For $x>0$, $y>0$ - For $x<0$, $y<0$ 5. **Limits:** - As $x \to +\infty$: $$\lim_{x \to +\infty} xe^{-x^2} = \lim_{x \to +\infty} \frac{x}{e^{x^2}} = 0$$ Because the exponential grows faster than any polynomial. - As $x \to -\infty$: $$\lim_{x \to -\infty} xe^{-x^2} = 0$$ For the same reason. - At $x=0$, $y=0$. 6. **First derivative:** Use the product rule: $$y = x \cdot e^{-x^2}$$ $$y' = 1 \cdot e^{-x^2} + x \cdot \frac{d}{dx} e^{-x^2}$$ Derivative of the exponential: $$\frac{d}{dx} e^{-x^2} = e^{-x^2} \cdot (-2x) = -2x e^{-x^2}$$ So: $$y' = e^{-x^2} - 2x^2 e^{-x^2} = e^{-x^2}(1 - 2x^2)$$ 7. **Critical points:** Set $y' = 0$: $$e^{-x^2}(1 - 2x^2) = 0 \implies 1 - 2x^2 = 0$$ $$2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$$ 8. **Second derivative:** Differentiate $y'$: $$y' = e^{-x^2}(1 - 2x^2)$$ Use product rule again: $$y'' = \frac{d}{dx} e^{-x^2} \cdot (1 - 2x^2) + e^{-x^2} \cdot \frac{d}{dx}(1 - 2x^2)$$ Calculate derivatives: $$\frac{d}{dx} e^{-x^2} = -2x e^{-x^2}$$ $$\frac{d}{dx}(1 - 2x^2) = -4x$$ So: $$y'' = (-2x e^{-x^2})(1 - 2x^2) + e^{-x^2}(-4x) = e^{-x^2}[-2x(1 - 2x^2) - 4x]$$ Simplify inside brackets: $$-2x + 4x^3 - 4x = 4x^3 - 6x$$ Therefore: $$y'' = e^{-x^2}(4x^3 - 6x) = 2x e^{-x^2}(2x^2 - 3)$$ 9. **Inflection points:** Set $y''=0$: $$2x e^{-x^2}(2x^2 - 3) = 0$$ Since $e^{-x^2} \neq 0$, solve: $$2x = 0 \implies x=0$$ $$2x^2 - 3 = 0 \implies x^2 = \frac{3}{2} \implies x = \pm \sqrt{\frac{3}{2}}$$ 10. **Summary:** - Domain: $(-\infty, +\infty)$ - Intercepts: $(0,0)$ - Positive for $x>0$, negative for $x<0$ - Limits at infinity: $0$ - Critical points at $x= \pm \frac{1}{\sqrt{2}}$ - Inflection points at $x=0$ and $x= \pm \sqrt{\frac{3}{2}}$ This completes the detailed analysis of the function $y = xe^{-x^2}$.