1. The problem is to find the Wendepunkt (inflection point) of the function $f(x) = 3x e^{-0.5x + 1}$.\n\n2. To find the inflection point, we need to find where the second derivative $f''(x)$ changes sign, i.e., where $f''(x) = 0$.\n\n3. First, find the first derivative $f'(x)$ using the product rule: $f(x) = u(x)v(x)$ with $u(x) = 3x$ and $v(x) = e^{-0.5x + 1}$.\n\n4. Compute $u'(x) = 3$ and $v'(x) = e^{-0.5x + 1} \cdot (-0.5) = -0.5 e^{-0.5x + 1}$.\n\n5. Then, $f'(x) = u'(x)v(x) + u(x)v'(x) = 3 e^{-0.5x + 1} + 3x (-0.5 e^{-0.5x + 1}) = 3 e^{-0.5x + 1} - 1.5 x e^{-0.5x + 1}$.\n\n6. Factor out $e^{-0.5x + 1}$: $$f'(x) = e^{-0.5x + 1} (3 - 1.5 x).$$\n\n7. Next, find the second derivative $f''(x)$ by differentiating $f'(x)$: $$f''(x) = \frac{d}{dx} \left(e^{-0.5x + 1} (3 - 1.5 x)\right).$$\n\n8. Use the product rule again with $p(x) = e^{-0.5x + 1}$ and $q(x) = 3 - 1.5 x$.\n\n9. Compute $p'(x) = -0.5 e^{-0.5x + 1}$ and $q'(x) = -1.5$.\n\n10. Then, $$f''(x) = p'(x) q(x) + p(x) q'(x) = -0.5 e^{-0.5x + 1} (3 - 1.5 x) + e^{-0.5x + 1} (-1.5).$$\n\n11. Factor out $e^{-0.5x + 1}$: $$f''(x) = e^{-0.5x + 1} \left(-0.5 (3 - 1.5 x) - 1.5\right).$$\n\n12. Simplify inside the parentheses: $$-0.5 (3 - 1.5 x) - 1.5 = -1.5 + 0.75 x - 1.5 = 0.75 x - 3.$$\n\n13. So, $$f''(x) = e^{-0.5x + 1} (0.75 x - 3).$$\n\n14. Since $e^{-0.5x + 1} > 0$ for all real $x$, the sign of $f''(x)$ depends on $0.75 x - 3$.\n\n15. Set $f''(x) = 0$ to find inflection points: $$0.75 x - 3 = 0 \Rightarrow 0.75 x = 3 \Rightarrow x = \frac{3}{0.75} = 4.$$\n\n16. Therefore, the Wendepunkt (inflection point) is at $x = 4$.\n\n17. To find the corresponding $y$ value, substitute $x=4$ into $f(x)$: $$f(4) = 3 \cdot 4 \cdot e^{-0.5 \cdot 4 + 1} = 12 e^{-2 + 1} = 12 e^{-1} = \frac{12}{e}.$$\n\n18. Final answer: The inflection point is at $$\boxed{\left(4, \frac{12}{e}\right)}.$$