The problem defines a function $f(x) = x + \frac{1}{x}$. It asks us to: 1. Determine the domain $D_f$ of $f$.

1. Problem Description

The problem defines a function $f(x) = x + \frac{1}{x}$. It asks us to:

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

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

3. Show that for distinct non-zero real numbers $a$ and $b$, $\frac{f(b)-f(a)}{b-a} = \frac{ba-1}{ba}$.

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

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

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

2. Solution Steps

1) Domain of $f$:

Since $f(x) = x + \frac{1}{x}$, $f$ is defined for all real numbers $x$ except when $x=0$.

Therefore, the domain of $f$ is $D_f = \mathbb{R} \setminus \{0\} = ]-\infty, 0[ \cup ]0, +\infty[$.

2) Show that $f$ is odd:

To show that $f$ is odd, we need to show that $f(-x) = -f(x)$ for all $x$ in the domain.

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

Thus, $f$ is an odd function.

3) Show that $\frac{f(b)-f(a)}{b-a} = \frac{ba-1}{ba}$:

$f(b) = b + \frac{1}{b}$ and $f(a) = a + \frac{1}{a}$.

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

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

4) Variations of $f$ on $[1, +\infty[$ and $]0, 1]$:

Let $x_1$ and $x_2$ be two real numbers such that $1 \le x_1 < x_2$. We have shown in part 3 that the rate of change between two distinct points is $\frac{x_1 x_2 - 1}{x_1 x_2}$. Since $x_1 \ge 1$ and $x_2 \ge 1$, $x_1 x_2 \ge 1$, so $x_1 x_2 - 1 \ge 0$. Since $x_1 > 0$ and $x_2 > 0$, $x_1 x_2 > 0$. Therefore, the ratio $\frac{x_1 x_2 - 1}{x_1 x_2} \ge 0$.

Since the rate of change between two values in the interval $[1, \infty[$ is positive, the function is increasing in this interval.

Let $0 < x_1 < x_2 < 1$. Then $0 < x_1 x_2 < 1$, so $x_1 x_2 - 1 < 0$. Since $x_1 > 0$ and $x_2 > 0$, we have $x_1 x_2 > 0$. Therefore, the ratio $\frac{x_1 x_2 - 1}{x_1 x_2} < 0$.

Since the rate of change between two values in the interval $]0, 1]$ is negative, the function is decreasing in this interval.

5) Variations of $f$ on $]-\infty, -1]$ and $[-1, 0[$:

Since $f$ is odd, if $1 \le x_1 < x_2$, then $-x_2 < -x_1 \le -1$. We know that $f(x_1) < f(x_2)$ on $[1, \infty[$.

Since $f(-x_2) = -f(x_2)$ and $f(-x_1) = -f(x_1)$, we have $-f(x_2) < -f(x_1)$. Thus $f(-x_2) < f(-x_1)$.

Therefore, if $-x_2 < -x_1 \le -1$, then $f(-x_2) < f(-x_1)$. So $f$ is increasing on $]-\infty, -1]$.

Similarly, if $0 < x_1 < x_2 < 1$, then $-1 < -x_2 < -x_1 < 0$. We know that $f(x_1) > f(x_2)$ on $]0, 1[$.

Since $f(-x_2) = -f(x_2)$ and $f(-x_1) = -f(x_1)$, we have $-f(x_2) > -f(x_1)$. Thus $f(-x_2) > f(-x_1)$.

Therefore, if $-1 < -x_2 < -x_1 < 0$, then $f(-x_2) > f(-x_1)$. So $f$ is decreasing on $[-1, 0[$.

6) Variation table of $f$ on $D_f$:

To construct the variation table, we need the critical points.

$f'(x) = 1 - \frac{1}{x^2}$. $f'(x) = 0$ when $1 = \frac{1}{x^2}$, which implies $x^2 = 1$, so $x = \pm 1$.

x | -inf -1 0 1 +inf

---------------------------------

f'(x)| + 0 - - 0 +

---------------------------------

f(x)| increasing -2 decreasing |decreasing 2 increasing

3. Final Answer

1) $D_f = ]-\infty, 0[ \cup ]0, +\infty[$

2) $f$ is odd.

3) $\frac{f(b)-f(a)}{b-a} = \frac{ba-1}{ba}$

4) $f$ is increasing on $[1, +\infty[$, $f$ is decreasing on $]0, 1]$.

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

6) See above table for the variation table of $f$ on $D_f$.