We are given the functions $f(x) = \frac{x^2}{5} + 5$ and $g(x) = \sqrt{5x - 25}$, along with their domains:
$M_f = [\sqrt{5}, \infty[$ and $M_g = [5, 10]$.

We need to determine the composite functions $f \circ g$, $g \circ f$, and $f \circ f$, along with their domains.

1. Composite Function $f \circ g$

The composite function $(f \circ g)(x)$ is defined as: $(f \circ g)(x) = f(g(x)).$ Substitute $g(x) = \sqrt{5x - 25}$ into $f(x)$: $f(g(x)) = \frac{(\sqrt{5x - 25})^2}{5} + 5 = \frac{5x - 25}{5} + 5 = x - 5 + 5 = x.$ Thus, $f(g(x)) = x$.

Domain of $f \circ g$:
The domain of $f \circ g$ is the set of $x$ values for which $g(x)$ is in the domain of $f(x)$. The domain of $g(x)$ is $[5, 10]$. The range of $g(x)$, which is $[0, \infty)$, is contained within the domain of $f(x)$. Therefore, the domain of $f \circ g$ is $[5, 10]$.

2. Composite Function $g \circ f$

The composite function $(g \circ f)(x)$ is defined as: $(g \circ f)(x) = g(f(x)).$ Substitute $f(x) = \frac{x^2}{5} + 5$ into $g(x)$: $g(f(x)) = \sqrt{5 \left( \frac{x^2}{5} + 5 \right) - 25} = \sqrt{x^2 + 25 - 25} = \sqrt{x^2}.$ Thus, $g(f(x)) = |x|$.

Domain of $g \circ f$:
The domain of $g \circ f$ is the set of $x$ values for which $f(x)$ is in the domain of $g(x)$. The domain of $f(x)$ is $[\sqrt{5}, \infty)$. We also need $5 \leq f(x) \leq 10$. Setting this condition on $f(x)$: $5 \leq \frac{x^2}{5} + 5 \leq 10.$ This simplifies to: $0 \leq \frac{x^2}{5} \leq 5 \quad \text{or} \quad 0 \leq x^2 \leq 25,$ which gives $0 \leq |x| \leq 5$. Therefore, $x \in [0, 5]$.

Thus, the domain of $g \circ f$ is $[\sqrt{5}, 5]$.

3. Composite Function $f \circ f$

The composite function $(f \circ f)(x)$ is defined as: $(f \circ f)(x) = f(f(x)).$ Substitute $f(x) = \frac{x^2}{5} + 5$ into itself: $f(f(x)) = \frac{\left( \frac{x^2}{5} + 5 \right)^2}{5} + 5.$ First, expand the expression: $\left( \frac{x^2}{5} + 5 \right)^2 = \frac{x^4}{25} + 2 \cdot \frac{x^2}{5} \cdot 5 + 25 = \frac{x^4}{25} + 2x^2 + 25.$ Now substitute this into $f(x)$: $f(f(x)) = \frac{\frac{x^4}{25} + 2x^2 + 25}{5} + 5 = \frac{x^4}{125} + \frac{2x^2}{5} + 5 + 5.$ Thus, $f(f(x)) = \frac{x^4}{125} + \frac{2x^2}{5} + 10.$

Domain of $f \circ f$:
The domain of $f \circ f$ is the set of $x$ values for which $f(x)$ is in the domain of $f(x)$, which is $[\sqrt{5}, \infty)$. Thus, the domain of $f \circ f$ is also $[\sqrt{5}, \infty)$.

Final Answer:

• $f \circ g(x) = x$, with domain $[5, 10]$.
• $g \circ f(x) = |x|$, with domain $[\sqrt{5}, 5]$.
• $f \circ f(x) = \frac{x^4}{125} + \frac{2x^2}{5} + 10$, with domain $[\sqrt{5}, \infty)$.

Would you like more details on any specific step or have any questions?

Here are 5 related questions to practice:

1. What happens to the domain when you switch the order of composition of functions?
2. How would the composition change if we altered $f(x)$ to $f(x) = x^2 + 1$?
3. Can you find the range of $f \circ g$ based on the given information?
4. What are the necessary conditions for two functions to be composable?
5. How would the composition change if we restrict the range of $g(x)$?

Tip: Always check the domain of the inside function in a composite function to ensure that it fits within the domain of the outside function.