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The image contains a math problem in Greek regarding two functions \( f(x) \) and \( g(x) \) with the following tasks:

Given functions:
- \( f(x) = \sqrt{x + 1} \)
- \( g(x) = 2 - x \)

You are asked to:

1. Define the composition function \( f \circ g \).
2. Prove that \( f \) is invertible and find \( f^{-1} \).
3. Determine the type of monotonicity (increasing or decreasing) of the function \( f \circ g \).

Let’s work through each part:

### i. Define the composition function \( f \circ g \):

The composition \( (f \circ g)(x) \) means applying \( g(x) \) first and then applying \( f \) to the result.

\[
(f \circ g)(x) = f(g(x)) = f(2 - x)
\]

Now, substitute \( g(x) = 2 - x \) into \( f(x) = \sqrt{x + 1} \):

\[
f(g(x)) = \sqrt{(2 - x) + 1} = \sqrt{3 - x}
\]

So the composition is:

\[
(f \circ g)(x) = \sqrt{3 - x}
\]

### ii. Prove that \( f(x) = \sqrt{x + 1} \) is invertible and find \( f^{-1}(x) \):

To prove that \( f \) is invertible, we need to show that it is one-to-one (injective) and onto (surjective).

1. **Injectivity**: 
   \( f(x) = \sqrt{x + 1} \) is an increasing function because its derivative is positive. The derivative of \( f(x) \) is:
   
   \[
   f'(x) = \frac{1}{2\sqrt{x+1}} > 0 \text{ for all } x \geq -1
   \]

   Since the derivative is always positive, \( f(x) \) is strictly increasing, which implies it is one-to-one.

2. **Finding the inverse**: 
   To find the inverse of \( f(x) = \sqrt{x + 1} \), solve for \( x \) in terms of \( y \):

   \[
   y = \sqrt{x + 1} \quad \Rightarrow \quad y^2 = x + 1 \quad \Rightarrow \quad x = y^2 - 1
   \]

   Therefore, the inverse function is:

   \[
   f^{-1}(x) = x^2 - 1
   \]

### iii. Determine the type of monotonicity of \( f \circ g \):

To find the monotonicity of \( (f \circ g)(x) = \sqrt{3 - x} \), we need to compute its derivative:

\[
(f \circ g)'(x) = \frac{d}{dx} \sqrt{3 - x} = \frac{-1}{2\sqrt{3 - x}}
\]

Since the derivative is negative for all \( x \), the function \( f \circ g \) is strictly decreasing.

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### Summary of Results:
1. \( (f \circ g)(x) = \sqrt{3 - x} \)
2. \( f^{-1}(x) = x^2 - 1 \)
3. \( (f \circ g)(x) \) is strictly decreasing.

Let me know if you would like further details or explanations.

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Here are 5 related questions for further practice:
1. Find the domain of \( f(x) \) and \( g(x) \) separately.
2. What is the domain of the composition \( f \circ g \)?
3. Calculate the range of \( f(x) \) and \( f \circ g(x) \).
4. Verify the inverse \( f^{-1}(x) \) by composing \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \).
5. Analyze the monotonicity of \( g(x) = 2 - x \) and compare it with that of \( f(x) \).

**Tip:** Always check the domain restrictions when working with square roots or inverse functions.

Δίνονται οι συναρτήσεις f(x) = √(x + 1) και g(x) = 2 - x. Να ορίσετε τη σύνθεση f ∘ g, αποδείξτε ότι η f είναι αντιστρέψιμη και βρείτε την f⁻¹, και να βρείτε το είδος της μονοτονίας της σύνθεσης f ∘ g.

Explore the composition of functions f(x) = √(x + 1) and g(x) = 2 - x, prove that f(x) is invertible, and find its inverse. This problem also investigates the monotonicity of the composition function. Suitable for students in Grades 10-12.

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