1. **State the problem:** We analyze the function $$f(x) = (2x + 1)e^{1 - 2x}$$ and find its derivative, limit behavior, tangent line at $$x_0 = \frac{1}{2}$$, and asymptotic properties. 2. **Recall the product rule for derivatives:** If $$f(x) = u(x)v(x)$$, then $$f'(x) = u'(x)v(x) + u(x)v'(x)$$. 3. **Calculate the derivative:** Let $$u(x) = 2x + 1$$ and $$v(x) = e^{1 - 2x}$$. - $$u'(x) = 2$$ - $$v'(x) = e^{1 - 2x} \times (-2) = -2e^{1 - 2x}$$ Using the product rule: $$f'(x) = 2e^{1 - 2x} + (2x + 1)(-2e^{1 - 2x}) = 2e^{1 - 2x} - 2(2x + 1)e^{1 - 2x}$$ Simplify: $$f'(x) = e^{1 - 2x} \left(2 - 4x - 2\right) = e^{1 - 2x}(-4x) = -4x e^{1 - 2x}$$ 4. **Evaluate the derivative at $$x_0 = \frac{1}{2}$$:** $$f'(\frac{1}{2}) = -4 \times \frac{1}{2} \times e^{1 - 2 \times \frac{1}{2}} = -2 e^{1 - 1} = -2 e^0 = -2$$ 5. **Find the function value at $$x_0$$:** $$f(\frac{1}{2}) = (2 \times \frac{1}{2} + 1) e^{1 - 2 \times \frac{1}{2}} = (1 + 1) e^{1 - 1} = 2 \times 1 = 2$$ 6. **Equation of the tangent line $$T$$ at $$x_0$$:** Using point-slope form: $$y = f(x_0) + f'(x_0)(x - x_0) = 2 - 2(x - \frac{1}{2}) = 2 - 2x + 1 = 3 - 2x$$ 7. **Limit behavior:** Given $$\lim_{x \to +\infty} x e^{-x} = 0$$, and since $$f(x) = (2x + 1) e^{1 - 2x} = e \times (2x + 1) e^{-2x}$$, as $$x \to +\infty$$, $$e^{-2x}$$ dominates and tends to zero faster than any polynomial grows, so: $$\lim_{x \to +\infty} f(x) = 0$$ 8. **Asymptote:** Since $$f(x) \to 0$$ as $$x \to +\infty$$, the x-axis $$y=0$$ is a horizontal asymptote. **Final answers:** - Derivative: $$f'(x) = -4x e^{1 - 2x}$$ - Tangent line at $$x_0 = \frac{1}{2}$$: $$y = 3 - 2x$$ - Horizontal asymptote: $$y = 0$$