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Let's go step by step through this exercise:

We are given the function \( f(x) = \frac{2x + 3}{x + 5} \), and the goal is to solve the following problems:

### 1. **Prove that \( f(x) \) can also be written as \( f(x) = 2 - \frac{7}{x + 5} \)**

We start by simplifying the original function:

\[
f(x) = \frac{2x + 3}{x + 5}
\]

We want to express this in the form \( f(x) = 2 - \frac{7}{x + 5} \). Let's rewrite the fraction:

\[
\frac{2x + 3}{x + 5} = \frac{(2x + 10) - 7}{x + 5} = \frac{2(x + 5) - 7}{x + 5}
\]

Now, we can split the fraction:

\[
\frac{2(x + 5) - 7}{x + 5} = \frac{2(x + 5)}{x + 5} - \frac{7}{x + 5} = 2 - \frac{7}{x + 5}
\]

So, we have shown that:

\[
f(x) = 2 - \frac{7}{x + 5}
\]

### 2. **Prove that \( f(x) \) is increasing on the interval \([-3, +\infty[\)**

To show that \( f(x) \) is increasing, we need to check the sign of the derivative \( f'(x) \).

Recall \( f(x) = \frac{2x + 3}{x + 5} \). Let's calculate the derivative using the quotient rule:

\[
f'(x) = \frac{(x + 5)(2) - (2x + 3)(1)}{(x + 5)^2}
\]

Simplifying the numerator:

\[
f'(x) = \frac{2(x + 5) - (2x + 3)}{(x + 5)^2} = \frac{2x + 10 - 2x - 3}{(x + 5)^2} = \frac{7}{(x + 5)^2}
\]

Since \( (x + 5)^2 \) is always positive for all \( x > -5 \), \( f'(x) = \frac{7}{(x + 5)^2} \) is positive for \( x \geq -3 \).

Thus, \( f(x) \) is increasing on the interval \([-3, +\infty[\).

### 3a. **Prove that \( f(x) \) admits a minimum and specify it**

Since \( f(x) \) is increasing, the minimum value of \( f(x) \) occurs at the left endpoint of the interval, which is \( x = -3 \).

Let's calculate \( f(-3) \):

\[
f(-3) = \frac{2(-3) + 3}{-3 + 5} = \frac{-6 + 3}{2} = \frac{-3}{2} = -\frac{3}{2}
\]

Therefore, the minimum value of \( f(x) \) is \( f(-3) = -\frac{3}{2} \).

### 3b. **Prove that \( f(x) \) admits an upper bound and specify it**

Since \( f(x) \) is increasing and unbounded as \( x \to +\infty \), there is no finite upper bound. However, for large values of \( x \), \( f(x) \) approaches 2:

\[
\lim_{x \to +\infty} f(x) = 2
\]

Thus, the function has an asymptote at \( y = 2 \), which can be considered its upper bound in the limit.

### 3c. **Deduce that \( f(x) \) is bounded and provide an interval containing \( f(x) \)**

From the previous results, we know that \( f(x) \) has a minimum of \( -\frac{3}{2} \) at \( x = -3 \) and approaches 2 as \( x \to +\infty \). Therefore, \( f(x) \) is bounded between \( -\frac{3}{2} \) and 2.

Thus, the interval containing \( f(x) \) is:

\[
f(x) \in \left[-\frac{3}{2}, 2\right[
\]

---

Would you like further details on any of the steps? Here are some related questions to expand your understanding:

1. What is the significance of an asymptote in a function like this?
2. Can you explain how the quotient rule for derivatives works?
3. How would the behavior of the function change if the denominator were different, say \( x + 3 \)?
4. How can you generalize this process for proving whether functions are increasing or decreasing?
5. Why does the function have a minimum, but not a maximum?

**Tip:** When working with increasing or decreasing functions, the derivative is a crucial tool. If the derivative is positive, the function is increasing, and if negative, it's decreasing!

Soit f la fonction définie sur [-3, +∞[ par : f(x) = (2x + 3) / (x + 5). 1) Démontrer que f(x) peut aussi s'écrire : f(x) = 2 - 7 / (x + 5). 2) Démontrer que f est croissante sur [-3, +∞[. 3) a) Démontrer que f admet un minimum, le préciser. b) Démontrer que f admet un majorant, en préciser un. c) En déduire que f est bornée et indiquer un encadrement de f(x).

This problem covers key topics like proving function equivalence, checking monotonicity using derivatives, and finding minima and asymptotes. Learn how to prove that f(x) = (2x + 3) / (x + 5) can also be written as 2 - 7 / (x + 5), analyze its increasing nature, and determine its boundedness.

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