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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.

Olkoot f(x) = \frac{x^2}{5} + 5 ja g(x) = \sqrt{5x - 25}, joille M_f = [\sqrt{5}, \infty[ ja M_g = [5, 10]. Määritä yhdistetyt funktiot f\circ g, g\circ f ja f\circ f määritysjoukkoineen.

Explore function compositions involving f(x) = x^2/5 + 5 and g(x) = sqrt(5x - 25), including the derivation of f(g(x)), g(f(x)), and f(f(x)) along with their domains. Suitable for high school students in Grades 11-12.

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