1. **Statement of the problem:** Given: - Curve (C) with equation $y = x^3 + 3x + 8$ - Line (\Delta) with equation $8 - 3x = y$ or $y = 8 - 3x$ Find the intersection point, study relative position, sign of $f(x)$, and solve inequalities. 2. **Intersection of (C) and (\Delta):** Set $x^3 + 3x + 8 = 8 - 3x$ Simplify: $$x^3 + 3x + 8 = 8 - 3x \implies x^3 + 3x + 3x = 0 \implies x^3 + 6x = 0$$ Factor: $$x(x^2 + 6) = 0$$ Roots: $$x=0$$ (real root), and $x^2 = -6$ (no real roots). Substitute $x=0$: $$y = 8 - 3(0) = 8$$ Intersection point: $(0, 8)$ 3. **Relative position of (\Delta) to (C):** - For $x=1$: $(C): y = 1^3 + 3*1 + 8 = 12$ $(\Delta): y = 8 - 3*1 = 5$ Since $12 > 5$, point of C lies above (\Delta) at $x=1$. - For $x=-1$: $(C): y = -1 + (-3) + 8 = 4$ $(\Delta): y = 8 - (-3) = 11$ Since $4 < 11$, point of C lies below (\Delta) at $x=-1$. Thus, (C) crosses (\Delta) at $(0,8)$, and the position shifts from below to above. 4. **Function $f(x) = x^3 + 3x + 8$, sign analysis:** - $f(x) = 0$ Check for approximate roots between $-1.6$ and $-1.5$: $$f(-1.6) = (-1.6)^3 + 3(-1.6) + 8 = -4.096 -4.8 + 8 = -0.896 < 0$$ $$f(-1.5) = (-1.5)^3 + 3(-1.5) + 8 = -3.375 - 4.5 + 8 = 0.125 > 0$$ So, sign changes between $-1.6$ and $-1.5$, confirming $$-1.6 < \alpha < -1.5$$ (the problem's statement about $-1.5 < \alpha < -1.6$ likely switches these) 5. **Function $f(x) = \frac{x^3 + 2x^2 - 2}{x^2 +1}$** Calculate limits: $$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \frac{x^3 + 2x^2 - 2}{x^2 + 1}$$ Divide numerator and denominator by $x^2$: $$= \lim_{x \to +\infty} \frac{x + 2 - 2/x^2}{1 + 1/x^2} = +\infty$$ Similarly, $$\lim_{x \to -\infty} f(x) = -\infty$$ 6. **Rewrite $f(x)$:** $$f(x) = x + 2 - \frac{4x + 1}{x^2 + 1}$$ 7. **Find oblique asymptote (d):** Oblique asymptote given by: $$y = x + 2$$ because the remainder term $-\frac{4x + 1}{x^2 + 1} \to 0$ as $x \to \pm \infty$. 8. **Derivative of $f(x)$:** Given: $$f'(x) = \frac{x · g(x)}{(x^2 + 1)^2}$$ for some $g(x)$ polynomial. 9. **Limit of difference quotient: ** $$\lim_{x \to \alpha} \frac{f(x) - f(\alpha)}{x - \alpha} = f'(\alpha)$$ Geometrically, it is the slope of the tangent line to $f$ at $x=\alpha$. 10. **Study variations of $f'$ and construct variation table:** Using critical points from zeros of $f'(x)$ and given $f(\alpha) \approx -0.3$. 11. **Plot (d) and (C_f), consider point $(0.84, 0)$ for $f(x) = 0$** 12. **Discuss number of solutions of $m + 1 = f(x)$ depending on $m$ (parameter):** 13. **Define $h(x) = [f(x)]^2$, compute derivative using chain rule:** $$h'(x) = 2 f(x) f'(x)$$ 14. **Construct variation table for $h(x)$ accordingly.**