Let's work through the exercises from the image step by step.

Exercice 1: Déterminer le domaine de définition des fonctions

1) $f(x) = 4x^2 + 10x - \frac{1}{x-5} + 2\pi$

For $f(x)$, the domain is all real numbers except where the function is undefined. The term $\frac{1}{x-5}$ causes a restriction since division by zero is not allowed. Therefore, the function is undefined when $x = 5$.

• Domain of $f(x)$: $\mathbb{R} \setminus \{5\}$.

2) $g(x) = \frac{2}{5x} - \frac{11}{2x+110}$

For $g(x)$, the two fractions introduce restrictions. The denominator in each fraction must not equal zero:

• $5x \neq 0 \Rightarrow x \neq 0$

• $2x + 110 \neq 0 \Rightarrow x \neq -55$

• Domain of $g(x)$: $\mathbb{R} \setminus \{0, -55\}$.

3) $h(x) = \sqrt{4x + 28}$

For $h(x)$, the expression inside the square root must be non-negative: $4x + 28 \geq 0 \Rightarrow x \geq -7$

• Domain of $h(x)$: $[-7, +\infty[$.

4) $k(x) = \sqrt{\frac{11 - 5x}{7 + 2x}}$

For $k(x)$, the expression inside the square root must be non-negative, and the denominator must not be zero: $\frac{11 - 5x}{7 + 2x} \geq 0 \quad \text{and} \quad 7 + 2x \neq 0 \Rightarrow x \neq -\frac{7}{2}$

We need to solve the inequality $\frac{11 - 5x}{7 + 2x} \geq 0$ by analyzing when the numerator and denominator have the same sign.

• Domain of $k(x)$: Solve the inequality and determine when the quotient is non-negative.

Exercice 2:

The function is $f(x) = 4x^2 + 24x + 66$.

1) Forme canonique de $f(x)$:

We complete the square to rewrite $f(x)$ in canonical form:

$f(x) = 4(x^2 + 6x) + 66$ $f(x) = 4\left((x+3)^2 - 9\right) + 66$ $f(x) = 4(x+3)^2 - 36 + 66$ $f(x) = 4(x+3)^2 + 30$

• Forme canonique: $f(x) = 4(x+3)^2 + 30$

2) Signe de $f(x)$:

Since $4(x+3)^2 \geq 0$ for all $x \in \mathbb{R}$, the minimum value of $f(x)$ is $30$ (achieved when $x = -3$).

• Signe de $f(x)$: $f(x) \geq 30$ for all $x \in \mathbb{R}$.

3) Résoudre l'inéquation $f(x) \leq 66$:

We solve $4(x+3)^2 + 30 \leq 66$: $4(x+3)^2 + 30 \leq 66$ $4(x+3)^2 \leq 36$ $(x+3)^2 \leq 9$ $-3 \leq x+3 \leq 3$ $-6 \leq x \leq 0$

• Solution: $x \in [-6, 0]$.

Would you like a more detailed explanation for any of these parts or need help with the domain of $k(x)$?

Additional Questions:

1. How do you complete the square for a general quadratic expression?
2. What are the steps to solve inequalities involving square roots like in $k(x)$?
3. Why is the minimum of $f(x) = 4(x+3)^2 + 30$ at $x = -3$?
4. Can you describe how the sign chart method is used in solving inequalities?
5. What additional constraints might arise when solving rational inequalities?

Tip:

Always check for domain restrictions before solving equations or inequalities, especially with square roots and fractions.