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Let's first break down the problem step-by-step to find the inverse of \( f \circ g \) based on the provided information.

### Functions given:
- \( f(x) = \arccos(2x + 1) \)
- \( g(x) = \frac{x + 0.5}{2x + 2} \)

We need to:
1. Find the expression for \( f \circ g(x) \), which means applying \( g(x) \) first and then applying \( f(x) \) to the result of \( g(x) \).
   
2. Once we have \( f \circ g(x) \), we'll determine the inverse of this composite function, \( (f \circ g)^{-1}(x) \).

### Step 1: Compute \( f \circ g(x) \)
The composite function \( f \circ g(x) \) is defined as:
\[
(f \circ g)(x) = f(g(x))
\]

Substituting \( g(x) = \frac{x + 0.5}{2x + 2} \) into \( f(x) = \arccos(2x + 1) \):

\[
f(g(x)) = \arccos\left( 2 \left( \frac{x + 0.5}{2x + 2} \right) + 1 \right)
\]

We simplify the expression inside the \( \arccos \):
\[
2 \cdot \frac{x + 0.5}{2x + 2} = \frac{2(x + 0.5)}{2x + 2} = \frac{2x + 1}{2x + 2}
\]

So, we now have:
\[
f(g(x)) = \arccos\left( \frac{2x + 1}{2x + 2} + 1 \right)
\]
Now, simplify the sum:
\[
f(g(x)) = \arccos\left( \frac{2x + 1 + 2x + 2}{2x + 2} \right) = \arccos\left( \frac{4x + 3}{2x + 2} \right)
\]

Thus, the expression for \( f \circ g(x) \) is:
\[
f \circ g(x) = \arccos\left( \frac{4x + 3}{2x + 2} \right)
\]

### Step 2: Find the inverse \( (f \circ g)^{-1}(x) \)
To find the inverse, we set \( y = f \circ g(x) \) and solve for \( x \) in terms of \( y \).

\[
y = \arccos\left( \frac{4x + 3}{2x + 2} \right)
\]
Apply \( \cos \) to both sides to eliminate the \( \arccos \):
\[
\cos(y) = \frac{4x + 3}{2x + 2}
\]

Now, solve this equation for \( x \):
\[
\cos(y)(2x + 2) = 4x + 3
\]
\[
2x\cos(y) + 2\cos(y) = 4x + 3
\]
Move all terms involving \( x \) to one side:
\[
2x\cos(y) - 4x = 3 - 2\cos(y)
\]
Factor out \( x \):
\[
x(2\cos(y) - 4) = 3 - 2\cos(y)
\]
Solve for \( x \):
\[
x = \frac{3 - 2\cos(y)}{2\cos(y) - 4}
\]

Thus, the inverse function of \( f \circ g(x) \) is:
\[
(f \circ g)^{-1}(x) = \frac{3 - 2\cos(x)}{2\cos(x) - 4}
\]

### Summary:
The inverse of the function \( f \circ g \) is given by:
\[
(f \circ g)^{-1}(x) = \frac{3 - 2\cos(x)}{2\cos(x) - 4}
\]

Would you like more details on the steps, or do you have any further questions? Here are five related questions for further exploration:

1. How do we find the domain of the function \( f \circ g(x) \)?
2. What are the specific domain restrictions for \( f(x) = \arccos(2x + 1) \)?
3. How does the derivative of \( f \circ g(x) \) behave? Is it increasing or decreasing?
4. Can \( f \circ g(x) \) be expressed as a simpler function using trigonometric identities?
5. What are the conditions under which the inverse function exists for \( f \circ g \)?

**Tip:** When working with composite functions, always apply the inner function first before proceeding with any other operations to avoid mistakes.

Given the function f(x) = arccos(2x + 1) and g(x) = (x + 0.5) / (2x + 2), find the inverse function of f ∘ g.

Explore the step-by-step solution to find the inverse of the composite function f ∘ g, where f(x) = arccos(2x + 1) and g(x) = (x + 0.5) / (2x + 2). Understand the use of composite functions, trigonometric identities, and inverse function properties in this advanced problem suitable for university-level students.

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