Solve in english

Given the function $y = f(x) = \frac{x}{(x-1)(x-2)}$.

i) The domain of definition of $f$.

Step 1: Identify restrictions for the function. The function is a rational function, so the denominator cannot be zero. $(x-1)(x-2) \ne 0$

Step 2: Solve for the values of $x$ that make the denominator zero. $x-1 = 0 \implies x = 1$ $x-2 = 0 \implies x = 2$

Step 3: State the domain. The domain of $f$ is all real numbers except $x=1$ and $x=2$. $D_f = \mathbb{R} \setminus \{1, 2\} = (-\infty, 1) \cup (1, 2) \cup (2, +\infty)$ The domain of definition of $f$ is $\boxed{\mathbb{R} \setminus \{1, 2\}}$.

ii) The limits at the bounds of the domain.

Step 1: Calculate the limits as $x \to \pm \infty$. $\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \frac{x}{(x-1)(x-2)} = \lim_{x \to \pm \infty} \frac{x}{x^2 - 3x + 2}$ Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^2$: $\lim_{x \to \pm \infty} \frac{\frac{1}{x}}{1 - \frac{3}{x} + \frac{2}{x^2}} = \frac{0}{1 - 0 + 0} = 0$ So, $\lim_{x \to -\infty} f(x) = \boxed{0}$ and $\lim_{x \to +\infty} f(x) = \boxed{0}$.

Step 2: Calculate the limits as $x \to 1$. As $x \to 1$, the numerator approaches $1$, and the denominator approaches $0$. This indicates a vertical asymptote. For $x \to 1^-$ (e.g., $x=0.9$): Numerator $x \to 1$ (positive). Denominator $(x-1) \to 0^-$ (negative). Denominator $(x-2) \to -1$ (negative). So, $\lim_{x \to 1^-} f(x) = \frac{1}{(0^-)(-1)} = \frac{1}{0^+} = \boxed{+\infty}$. For $x \to 1^+$ (e.g., $x=1.1$): Numerator $x \to 1$ (positive). Denominator $(x-1) \to 0^+$ (positive). Denominator $(x-2) \to -1$ (negative). So, $\lim_{x \to 1^+} f(x) = \frac{1}{(0^+)(-1)} = \frac{1}{0^-} = \boxed{-\infty}$.

Step 3: Calculate the limits as $x \to 2$. As $x \to 2$, the numerator approaches $2$, and the denominator approaches $0$. This indicates a vertical asymptote. For $x \to 2^-$ (e.g., $x=1.9$): Numerator $x \to 2$ (positive). Denominator $(x-1) \to 1$ (positive). Denominator $(x-2) \to 0^-$ (negative). So, $\lim_{x \to 2^-} f(x) = \frac{2}{(1)(0^-)} = \frac{2}{0^-} = \boxed{-\infty}$. For $x \to 2^+$ (e.g., $x=2.1$): Numerator $x \to 2$ (positive). Denominator $(x-1) \to 1$ (positive). Denominator $(x-2) \to 0^+$ (positive). So, $\lim_{x \to 2^+} f(x) = \frac{2}{(1)(0^+)} = \frac{2}{0^+} = \boxed{+\infty}$.

iii) The asymptotes, stationary points, the variation table and sketch the graph of the function $y=f(x)$.

Step 1: Determine the asymptotes. From the limits:

• Vertical asymptotes: $x=1$ and $x=2$.
• Horizontal asymptote: $y=0$ (as $x \to \pm \infty$). There are no oblique asymptotes.

Step 2: Find the stationary points. First, find the derivative $f'(x)$ using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$. Let $u = x \implies u' = 1$. Let $v = (x-1)(x-2) = x^2 - 3x + 2 \implies v' = 2x - 3$. $f'(x) = \frac{1 \cdot (x^2 - 3x + 2) - x \cdot (2x - 3)}{(x^2 - 3x + 2)^2}$ $f'(x) = \frac{x^2 - 3x + 2 - 2x^2 + 3x}{(x^2 - 3x + 2)^2}$ $f'(x) = \frac{-x^2 + 2}{(x^2 - 3x + 2)^2}$ Set $f'(x) = 0$ to find critical points: $-x^2 + 2 = 0 \implies x^2 = 2 \implies x = \pm \sqrt{2}$ Now, find the corresponding $y$-values: For $x = \sqrt{2}$: $f(\sqrt{2}) = \frac{\sqrt{2}}{(\sqrt{2}-1)(\sqrt{2}-2)} = \frac{\sqrt{2}}{2 - 2\sqrt{2} - \sqrt{2} + 2} = \frac{\sqrt{2}}{4 - 3\sqrt{2}}$ Rationalize the denominator: $f(\sqrt{2}) = \frac{\sqrt{2}(4 + 3\sqrt{2})}{(4 - 3\sqrt{2})(4 + 3\sqrt{2})} = \frac{4\sqrt{2} + 3(2)}{16 - 9(2)} = \frac{4\sqrt{2} + 6}{16 - 18} = \frac{4\sqrt{2} + 6}{-2} = -2\sqrt{2} - 3$ So, one stationary point is $(\sqrt{2}, -2\sqrt{2}-3) \approx (1.41, -5.83)$.

For $x = -\sqrt{2}$: $f(-\sqrt{2}) = \frac{-\sqrt{2}}{(-\sqrt{2}-1)(-\sqrt{2}-2)} = \frac{-\sqrt{2}}{(\sqrt{2}+1)(\sqrt{2}+2)} = \frac{-\sqrt{2}}{2 + 2\sqrt{2} + \sqrt{2} + 2} = \frac{-\sqrt{2}}{4 + 3\sqrt{2}}$ Rationalize the denominator: $f(-\sqrt{2}) = \frac{-\sqrt{2}(4 - 3\sqrt{2})}{(4 + 3\sqrt{2})(4 - 3\sqrt{2})} = \frac{-4\sqrt{2} + 3(2)}{16 - 9(2)} = \frac{-4\sqrt{2} + 6}{16 - 18} = \frac{-4\sqrt{2} + 6}{-2} = 2\sqrt{2} - 3$ So, the other stationary point is $(-\sqrt{2}, 2\sqrt{2}-3) \approx (-1.41, -0.17)$.

Step 3: Construct the variation table. The sign of $f'(x) = \frac{-x^2 + 2}{(x^2 - 3x + 2)^2}$ is determined by the numerator $-x^2 + 2$, as the denominator is always positive for $x \ne 1, 2$. $-x^2 + 2 > 0 \implies x^2 < 2 \implies -\sqrt{2} < x < \sqrt{2}$. $-x^2 + 2 < 0 \implies x^2 > 2 \implies x < -\sqrt{2}$ or $x > \sqrt{2}$. The critical points are $-\sqrt{2} \approx -1.41$, $1$, $\sqrt{2} \approx 1.41$, $2$.

Step 4: Sketch the graph. Based on the analysis:

• The graph approaches the horizontal asymptote $y=0$ from above as $x \to -\infty$.
• It decreases to a local minimum at $(-\sqrt{2}, 2\sqrt{2}-3) \approx (-1.41, -0.17)$.
• It then increases, crossing the x-axis at $x=0$ (since $f(0)=0$), and approaches $+\infty$ as $x \to 1^-$.
• Between $x=1$ and $x=2$, the graph starts from $-\infty$ as $x \to 1^+$.
• It increases to a local maximum at $(\sqrt{2}, -2\sqrt{2}-3) \approx (1.41, -5.83)$.
• It then decreases and approaches $-\infty$ as $x \to 2^-$.
• For $x > 2$, the graph starts from $+\infty$ as $x \to 2^+$.
• It decreases and approaches the horizontal asymptote $y=0$ from above as $x \to +\infty$.

(A visual sketch cannot be provided in this text format, but the description above outlines the shape and key points for plotting.)