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!