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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.

Determine the domain of definition for the given functions and solve quadratic inequalities.

Explore how to determine the domain of functions and solve quadratic inequalities, including square roots and rational expressions. This step-by-step guide is ideal for students in grades 10-12.

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