1. **Statement of the problem:** We are given that the curve $(C_f)$ of the function $f(x) = (x+1)e^{-x} - x$ intersects the x-axis at two points with abscissas $x_1$ and $x_2$ such that $-1.4 < x_1 < -1.3$ and $0.7 < x_2 < 0.8$. 2. **Goal:** Construct the line $(\Delta)$ with equation $y = -x$, the tangent line $(T)$ to $(C_f)$ at $x=0$, and the curve $(C_f)$ itself on the domain $[-2, +\infty[$. 3. **Recall the functions:** - $f(x) = (x+1)e^{-x} - x$ - $\Delta: y = -x$ - Tangent $T$ at $x=0$ has equation $y = f'(0)(x-0) + f(0)$ 4. **Calculate $f(0)$:** $$f(0) = (0+1)e^{0} - 0 = 1 \times 1 - 0 = 1$$ 5. **Calculate $f'(x)$:** Given in the problem: $f'(x) = -x g(x)$ where $g(x) = e^{-x} + \frac{1}{x}$. Calculate $f'(0)$ by taking the limit as $x \to 0$: - $g(x) = e^{-x} + \frac{1}{x}$ is not defined at 0, but we can find $\lim_{x \to 0} f'(x)$. Since $f'(x) = -x g(x) = -x \left(e^{-x} + \frac{1}{x}\right) = -x e^{-x} - 1$, then $$\lim_{x \to 0} f'(x) = \lim_{x \to 0} (-x e^{-x} - 1) = 0 - 1 = -1$$ So $f'(0) = -1$. 6. **Equation of tangent $T$ at $x=0$:** $$y = f'(0)(x - 0) + f(0) = -1 \times x + 1 = -x + 1$$ 7. **Summary:** - Line $\Delta$: $y = -x$ - Tangent $T$: $y = -x + 1$ - Curve $C_f$: $f(x) = (x+1)e^{-x} - x$ 8. **Domain for plotting:** $[-2, +\infty[$ This completes the construction of $(\Delta)$, $(T)$, and $(C_f)$ on the given domain.