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.