1. **Statement of the problem:** We have the function $$f(x) = \frac{1}{6}x^3 - x^2$$ defined on $$\mathbb{R}$$. We want to find: - $$\lim_{x \to +\infty} f(x)$$ and $$\lim_{x \to -\infty} f(x)$$. - Study the monotonicity of $$f$$ using its derivative $$f'(x) = \frac{1}{2}x(x-4)$$. - Show that the curve $$C_f$$ has an inflection point $$A$$ and find its coordinates. - Show that the line $$y = -2x + \frac{4}{3}$$ is tangent to $$C_f$$ at the point where $$x=2$$. - Solve $$f(x) = 0$$. - Calculate $$f(-1)$$ and $$f(7)$$. 2. **Limits at infinity:** Since $$f(x) = \frac{1}{6}x^3 - x^2$$, the dominant term as $$x \to \pm \infty$$ is $$\frac{1}{6}x^3$$. - As $$x \to +\infty$$, $$x^3 \to +\infty$$, so $$f(x) \to +\infty$$. - As $$x \to -\infty$$, $$x^3 \to -\infty$$, so $$f(x) \to -\infty$$. 3. **Monotonicity study:** Given $$f'(x) = \frac{1}{2}x(x-4)$$. - Find critical points by solving $$f'(x) = 0$$: $$\frac{1}{2}x(x-4) = 0 \implies x=0 \text{ or } x=4$$. - Sign of $$f'(x)$$: - For $$x < 0$$, $$x < 0$$ and $$x-4 < 0$$, so $$f'(x) = \frac{1}{2} \times (negative) \times (negative) = positive$$. - For $$0 < x < 4$$, $$x > 0$$ but $$x-4 < 0$$, so $$f'(x) = \frac{1}{2} \times (positive) \times (negative) = negative$$. - For $$x > 4$$, both $$x > 0$$ and $$x-4 > 0$$, so $$f'(x) = \frac{1}{2} \times (positive) \times (positive) = positive$$. - Therefore, $$f$$ is increasing on $$(-\infty, 0)$$, decreasing on $$(0,4)$$, and increasing on $$(4, +\infty)$$. - **Table of variations:** | Interval | $$(-\infty, 0)$$ | $$0$$ | $$(0,4)$$ | $$4$$ | $$(4, +\infty)$$ | |----------|------------------|-------|----------|-------|-----------------| | $$f'(x)$$ | $$+$$ | $$0$$ | $$-$$ | $$0$$ | $$+$$ | | $$f(x)$$ | Increasing | Max | Decreasing| Min | Increasing | 4. **Inflection point:** - The inflection point occurs where $$f''(x) = 0$$. - Compute $$f''(x)$$: $$f'(x) = \frac{1}{2}x(x-4) = \frac{1}{2}(x^2 - 4x)$$ $$f''(x) = \frac{1}{2}(2x - 4) = x - 2$$. - Solve $$f''(x) = 0$$: $$x - 2 = 0 \implies x = 2$$. - Find $$y$$ coordinate of inflection point $$A$$: $$f(2) = \frac{1}{6} \times 2^3 - 2^2 = \frac{1}{6} \times 8 - 4 = \frac{8}{6} - 4 = \frac{4}{3} - 4 = -\frac{8}{3}$$. - So, $$A(2, -\frac{8}{3})$$. 5. **Tangent line at $$x=2$$:** - Slope of tangent is $$f'(2)$$: $$f'(2) = \frac{1}{2} \times 2 \times (2-4) = \frac{1}{2} \times 2 \times (-2) = -2$$. - Equation of tangent line at $$x=2$$: $$y = f(2) + f'(2)(x - 2) = -\frac{8}{3} - 2(x - 2) = -2x + 4 - \frac{8}{3} = -2x + \frac{4}{3}$$. - This matches the given line $$y = -2x + \frac{4}{3}$$, confirming it is tangent at $$x=2$$. 6. **Solve $$f(x) = 0$$:** $$\frac{1}{6}x^3 - x^2 = 0 \implies x^2 \left( \frac{1}{6}x - 1 \right) = 0$$ - So, $$x^2 = 0 \implies x=0$$ - Or $$\frac{1}{6}x - 1 = 0 \implies x = 6$$ - Solutions: $$x=0$$ and $$x=6$$. 7. **Calculate $$f(-1)$$ and $$f(7)$$:** - $$f(-1) = \frac{1}{6}(-1)^3 - (-1)^2 = \frac{1}{6}(-1) - 1 = -\frac{1}{6} - 1 = -\frac{7}{6}$$. - $$f(7) = \frac{1}{6} \times 7^3 - 7^2 = \frac{1}{6} \times 343 - 49 = \frac{343}{6} - 49 = \frac{343 - 294}{6} = \frac{49}{6}$$. **Final answers:** - $$\lim_{x \to +\infty} f(x) = +\infty$$ - $$\lim_{x \to -\infty} f(x) = -\infty$$ - $$f$$ increases on $$(-\infty, 0)$$, decreases on $$(0,4)$$, increases on $$(4, +\infty)$$. - Inflection point $$A(2, -\frac{8}{3})$$. - Tangent line at $$x=2$$ is $$y = -2x + \frac{4}{3}$$. - Roots of $$f(x) = 0$$ are $$x=0$$ and $$x=6$$. - $$f(-1) = -\frac{7}{6}$$ and $$f(7) = \frac{49}{6}$$.