We are given the function $f(x) = \frac{x}{x^2 - 1}$. We need to: 1. Determine the domain $D_f$ of the function $f$.

1. Problem Description

We are given the function $f(x) = \frac{x}{x^2 - 1}$. We need to:

1. Determine the domain $D_f$ of the function $f$.

2. Verify that $f$ is an odd function.

3. Show that for all distinct real numbers $a$ and $b$ in $D_f$, $T = \frac{f(a) - f(b)}{a - b} = -\frac{ab + 1}{(a^2 - 1)(b^2 - 1)}$.

4. Study the variations of $f$ on the intervals $[0, 1[$ and $]1, +\infty[$.

5. Deduce the variations of $f$ on the intervals $]-1, 0]$ and $]-\infty, -1[$.

6. Draw the variation table of $f$ on $D_f$.

2. Solution Steps

1. Determine the domain $D_f$ of $f(x) = \frac{x}{x^2 - 1}$.

The function is defined when the denominator is not zero.

$x^2 - 1 \neq 0 \implies x^2 \neq 1 \implies x \neq \pm 1$.

Therefore, $D_f = \mathbb{R} \setminus \{-1, 1\} = ]-\infty, -1[ \cup ]-1, 1[ \cup ]1, +\infty[$.

2. Verify that $f$ is an odd function.

A function is odd if $f(-x) = -f(x)$ for all $x$ in its domain.

$f(-x) = \frac{-x}{(-x)^2 - 1} = \frac{-x}{x^2 - 1} = -\frac{x}{x^2 - 1} = -f(x)$.

Thus, $f$ is an odd function.

3. Show that $T = \frac{f(a) - f(b)}{a - b} = -\frac{ab + 1}{(a^2 - 1)(b^2 - 1)}$.

$f(a) = \frac{a}{a^2 - 1}$ and $f(b) = \frac{b}{b^2 - 1}$.

$f(a) - f(b) = \frac{a}{a^2 - 1} - \frac{b}{b^2 - 1} = \frac{a(b^2 - 1) - b(a^2 - 1)}{(a^2 - 1)(b^2 - 1)} = \frac{ab^2 - a - ba^2 + b}{(a^2 - 1)(b^2 - 1)} = \frac{ab^2 - ba^2 - a + b}{(a^2 - 1)(b^2 - 1)} = \frac{ab(b - a) + (b - a)}{(a^2 - 1)(b^2 - 1)} = \frac{(b - a)(ab + 1)}{(a^2 - 1)(b^2 - 1)} = -\frac{(a - b)(ab + 1)}{(a^2 - 1)(b^2 - 1)}$.

Thus, $T = \frac{f(a) - f(b)}{a - b} = \frac{-\frac{(a - b)(ab + 1)}{(a^2 - 1)(b^2 - 1)}}{a - b} = -\frac{ab + 1}{(a^2 - 1)(b^2 - 1)}$.

4. Study the variations of $f$ on $[0, 1[$ and $]1, +\infty[$.

Let $a, b \in [0, 1[$ such that $a < b$. Then $ab+1 > 0$. Also, $a^2 - 1 < 0$ and $b^2 - 1 < 0$, so $(a^2 - 1)(b^2 - 1) > 0$. Thus, $T = -\frac{ab+1}{(a^2-1)(b^2-1)} < 0$. Since $a<b$ and $T<0$, $f(a) > f(b)$. Therefore, $f$ is strictly decreasing on $[0, 1[$.

Let $a, b \in ]1, +\infty[$ such that $a < b$. Then $ab+1 > 0$. Also, $a^2 - 1 > 0$ and $b^2 - 1 > 0$, so $(a^2 - 1)(b^2 - 1) > 0$. Thus, $T = -\frac{ab+1}{(a^2-1)(b^2-1)} < 0$. Since $a<b$ and $T<0$, $f(a) > f(b)$. Therefore, $f$ is strictly decreasing on $]1, +\infty[$.

5. Deduce the variations of $f$ on $]-1, 0]$ and $]-\infty, -1[$.

Since $f$ is an odd function, $f(-x) = -f(x)$.

If $x \in [0, 1[$, then $-x \in ]-1, 0]$. $f$ is decreasing on $[0, 1[$ means that for $0 \le a < b < 1$, we have $f(a) > f(b)$. Thus, $-f(a) < -f(b)$. Since $f$ is odd, $f(-a) < f(-b)$. But $-b < -a$, so for $-b < -a \le 0$, we have $f(-b) > f(-a)$. This means $f$ is decreasing on $]-1, 0]$.

If $x \in ]1, \infty[$, then $-x \in ]-\infty, -1[$. $f$ is decreasing on $]1, \infty[$ means that for $1 < a < b$, we have $f(a) > f(b)$. Thus, $-f(a) < -f(b)$. Since $f$ is odd, $f(-a) < f(-b)$. But $-b < -a$, so for $-\infty < -b < -a < -1$, we have $f(-b) > f(-a)$. This means $f$ is decreasing on $]-\infty, -1[$.

6. Draw the variation table of $f$ on $D_f$.

| x | -inf | -1 | | 0 | | 1 | +inf |

| -------- | ---- | -- | ------- | --- | ----- | -- | ---- |

| f'(x) | | | - | | - | | |

| f(x) | 0 | || | 0 | | || | 0 |

On $]-\infty, -1[$, the function is decreasing from 0 to $-\infty$.

On $]-1, 0]$, the function is decreasing from $+\infty$ to

0. On $[0, 1[$, the function is decreasing from 0 to $-\infty$.

On $]1, +\infty[$, the function is decreasing from $+\infty$ to

0.

3. Final Answer

1. $D_f = ]-\infty, -1[ \cup ]-1, 1[ \cup ]1, +\infty[$

2. $f$ is an odd function.

3. $T = -\frac{ab + 1}{(a^2 - 1)(b^2 - 1)}$

4. $f$ is strictly decreasing on $[0, 1[$ and $]1, +\infty[$.

5. $f$ is strictly decreasing on $]-1, 0]$ and $]-\infty, -1[$.

6. Variation table:

| x | -inf | -1 | | 0 | | 1 | +inf |

| -------- | ---- | -- | ------- | --- | ----- | -- | ---- |

| f'(x) | | | - | | - | | |

| f(x) | 0 | || | 0 | | || | 0 |