The problem involves analyzing the function $f(x) = \frac{x}{x^2-1}$. We need to determine the domain, check if it's an odd function, prove a given equality, study the variations on specified intervals, deduce the variations on other intervals, and finally, create the variation table.

1. Problem Description

The problem involves analyzing the function $f(x) = \frac{x}{x^2-1}$. We need to determine the domain, check if it's an odd function, prove a given equality, study the variations on specified intervals, deduce the variations on other intervals, and finally, create the variation table.

2. Solution Steps

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

The function $f(x) = \frac{x}{x^2-1}$ is defined when the denominator is not zero.

$x^2 - 1 \neq 0$

$x^2 \neq 1$

$x \neq \pm 1$

Therefore, the domain is $D_f = \mathbb{R} \setminus \{-1, 1\} = (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.

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

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

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

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

3) Show that for all real numbers $a$ and $b$ distinct from $D_f$, we have:

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

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

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

Then,

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

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

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

Let $a, b \in ]1, \infty[$ with $a \neq b$.

Since $a > 1$ and $b > 1$, we have $a^2 > 1$ and $b^2 > 1$, so $a^2 - 1 > 0$ and $b^2 - 1 > 0$. Also $ab > 1$, so $ab + 1 > 0$.

Thus, $T = -\frac{ab + 1}{(a^2 - 1)(b^2 - 1)} < 0$.

Since $T < 0$, the function $f$ is decreasing on $]1, \infty[$.

Let $a, b \in [0, 1[$ with $a \neq b$.

Since $0 \le a < 1$ and $0 \le b < 1$, we have $a^2 < 1$ and $b^2 < 1$, so $a^2 - 1 < 0$ and $b^2 - 1 < 0$. Also $0 \le ab < 1$, so $1 \le ab + 1 < 2$.

Thus, $T = -\frac{ab + 1}{(a^2 - 1)(b^2 - 1)} = -\frac{(+)}{(-)(-)} = -\frac{(+)}{(+)} < 0$.

Since $T < 0$, the function $f$ is decreasing on $[0, 1[$.

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

Since $f$ is odd, $f(-x) = -f(x)$. If $x$ increases, then $-x$ decreases.

If $f$ is decreasing on $[0, 1[$, then it is also decreasing on $]-1, 0]$.

If $f$ is decreasing on $]1, \infty[$, then it is also 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 || -inf | | 0 | | +inf|| 0

3. Final Answer

1) $D_f = (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$

2) $f$ is odd

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

4) $f$ is decreasing on $[0, 1[$ and $]1, \infty[$

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

6) Variation table as described above.