Given the function $f(x) = \frac{x^2+3}{x+1}$, we need to: 1. Determine the domain of definition of $f$.

1. Problem Description

Given the function $f(x) = \frac{x^2+3}{x+1}$, we need to:

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

2. Calculate the limits of $f$ at the boundaries of its domain.

3. Show that for all $x \in ]-\infty, -1[ \cup ]-1, +\infty[$, $f'(x) = \frac{x^2+2x-3}{(x+1)^2}$.

4. Study the sense of variation of $f$ and draw its variation table.

5. Determine the real numbers $a$, $b$, and $c$ such that $f(x) = ax + b + \frac{c}{x+1}$.

6. Given that $a=1$, $b=-1$ and $c=4$, show that the line $(\Delta)$ with equation $y=x-1$ is an oblique asymptote to the curve $(C)$ at $-\infty$ and $+\infty$.

7. Draw the curve $(C)$ and the line $(\Delta)$ in the same coordinate system.

8. Let $g$ be the restriction of $f$ to the interval $I = ]-1, +\infty[$ such that $g(x)=f(-x)$.

9. Without studying the function $g$, draw its variation table.

1

0. Draw the curve $(C')$ of $g$ in the same coordinate system as $(C)$.

2. Solution Steps

1. Domain of definition:

The function $f(x) = \frac{x^2+3}{x+1}$ is defined for all $x$ such that the denominator is not zero. Thus, $x+1 \ne 0$, which means $x \ne -1$.

Therefore, the domain of definition is $D_f = \mathbb{R} \setminus \{-1\} = ]-\infty, -1[ \cup ]-1, +\infty[$.

2. Limits at the boundaries of the domain:

$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{x^2+3}{x+1} = \lim_{x \to -\infty} \frac{x^2}{x} = \lim_{x \to -\infty} x = -\infty$

$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \frac{x^2+3}{x+1} = \lim_{x \to +\infty} \frac{x^2}{x} = \lim_{x \to +\infty} x = +\infty$

$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{x^2+3}{x+1} = \frac{(-1)^2+3}{-1^-+1} = \frac{4}{0^-} = -\infty$

$\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{x^2+3}{x+1} = \frac{(-1)^2+3}{-1^++1} = \frac{4}{0^+} = +\infty$

3. Derivative:

$f(x) = \frac{x^2+3}{x+1}$

$f'(x) = \frac{(2x)(x+1) - (x^2+3)(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2 - 3}{(x+1)^2} = \frac{x^2+2x-3}{(x+1)^2}$

4. Sense of variation:

We study the sign of $f'(x) = \frac{x^2+2x-3}{(x+1)^2}$.

Since $(x+1)^2 > 0$ for $x \ne -1$, the sign of $f'(x)$ depends on the sign of $x^2+2x-3$.

$x^2+2x-3 = 0 \Rightarrow x = \frac{-2 \pm \sqrt{4 - 4(-3)}}{2} = \frac{-2 \pm \sqrt{16}}{2} = \frac{-2 \pm 4}{2}$. Thus $x_1 = 1$ and $x_2 = -3$.

The quadratic $x^2+2x-3$ is positive for $x \in ]-\infty, -3[ \cup ]1, +\infty[$ and negative for $x \in ]-3, 1[$.

So $f'(x) > 0$ for $x \in ]-\infty, -3[ \cup ]1, +\infty[$ and $f'(x) < 0$ for $x \in ]-3, -1[ \cup ]-1, 1[$.

Therefore, $f$ is increasing on $]-\infty, -3]$ and $[1, +\infty[$, and decreasing on $[-3, -1[$ and $]-1, 1]$.

5. Decomposition:

$f(x) = \frac{x^2+3}{x+1} = \frac{x^2-1+4}{x+1} = \frac{(x-1)(x+1)+4}{x+1} = x-1 + \frac{4}{x+1}$.

So $a=1, b=-1, c=4$.

6. Oblique asymptote:

$f(x) - (x-1) = \frac{4}{x+1}$.

$\lim_{x \to \pm \infty} [f(x) - (x-1)] = \lim_{x \to \pm \infty} \frac{4}{x+1} = 0$.

Thus, the line $y=x-1$ is an oblique asymptote to the curve $(C)$ at $-\infty$ and $+\infty$.

7. Drawing the curve and the line.

8. Restriction $g(x)=f(-x)$:

$g(x) = f(-x) = \frac{(-x)^2+3}{-x+1} = \frac{x^2+3}{1-x}$, with $x \in ]-1, +\infty[$. Thus, $-x \in ]-\infty, 1[$.

However, the domain is $]-1, +\infty[$. Since $g(x) = f(-x)$, then $g'(x) = -f'(-x)$.

If $x \in ]-1, +\infty[$, then $-x \in ]-\infty, 1[$.

When $f'(-x) > 0$, $-x \in ]-\infty, -3[ \cup ]1, +\infty[$, which implies $x \in ]-\infty, -1[ \cup ]3, +\infty[$.

Since the domain is $]-1, +\infty[$, $-x$ should be in $]-\infty, 1[$, thus $g'(x) = - f'(-x)$ should be examined when the given interval is $]-1, +\infty[$.

Also, $g'(x)= -f'(-x) = - \frac{(-x)^2+2(-x)-3}{(-x+1)^2}= - \frac{x^2-2x-3}{(1-x)^2}$.

The sign is determined by the numerator now. The roots are $x=3, x=-1$. $]-1, 3[$ is decreasing and $]3, +\infty[$ is increasing.

Therefore, $g$ is increasing when $x \in ]-1, 3[$ and decreasing when $x \in ]3, +\infty[$.

9. Variation table of $g$:

$g(x)$ increases on $]-1, 3[$ and decreases on $]3, +\infty[$. $g(3) = \frac{12}{-2}=-6$.

$\lim_{x \to -1^+} g(x) = \frac{4}{2} = 2$

$\lim_{x \to +\infty} g(x) = \lim_{x \to +\infty} \frac{x^2+3}{1-x} = -\infty$

1

0. Drawing the curve $(C')$.

3. Final Answer

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

2. $\lim_{x \to -\infty} f(x) = -\infty$, $\lim_{x \to +\infty} f(x) = +\infty$, $\lim_{x \to -1^-} f(x) = -\infty$, $\lim_{x \to -1^+} f(x) = +\infty$

3. $f'(x) = \frac{x^2+2x-3}{(x+1)^2}$

4. $f$ is increasing on $]-\infty, -3]$ and $[1, +\infty[$, and decreasing on $[-3, -1[$ and $]-1, 1]$.

5. $a=1, b=-1, c=4$

6. $y=x-1$ is an oblique asymptote to the curve $(C)$ at $-\infty$ and $+\infty$.

7. Graph of the curve and line.

8. $g(x)=f(-x) = \frac{x^2+3}{1-x}$,

9. $g$ is increasing when $x \in ]-1, 3[$ and decreasing when $x \in ]3, +\infty[$.

1

0. Graph of the curve $(C')$.