To solve this problem, we need to break it down step by step by focusing on the operations related to the composition of inverse functions.

We are given the following functions:

• $f(x) = x + 5$
• $g(x) = 4x - 3$
• $h(x) = 2x + 1$

Step 1: Find the inverse of each function

To find the inverse of a function, we reverse the operations in the original function. This means solving for $x$ in terms of $y$ when $y = f(x), g(x), h(x)$.

Inverse of $f(x) = x + 5$:

Let $y = f(x)$. This gives us: $y = x + 5$ Solving for $x$: $x = y - 5$ Thus, $f^{-1}(x) = x - 5$.

Inverse of $g(x) = 4x - 3$:

Let $y = g(x)$. This gives us: $y = 4x - 3$ Solving for $x$: $x = \frac{y + 3}{4}$ Thus, $g^{-1}(x) = \frac{x + 3}{4}$.

Inverse of $h(x) = 2x + 1$:

Let $y = h(x)$. This gives us: $y = 2x + 1$ Solving for $x$: $x = \frac{y - 1}{2}$ Thus, $h^{-1}(x) = \frac{x - 1}{2}$.

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Step 2: Verify the given options

Now we will check the composition of inverse functions as described in the options.

A. $(f^{-1} \circ g^{-1} \circ h^{-1})(x) = \frac{x + 45}{8}$

We need to compute $f^{-1}(g^{-1}(h^{-1}(x)))$:

1. Start with $h^{-1}(x) = \frac{x - 1}{2}$.
2. Next, apply $g^{-1}$ to this: $g^{-1}(h^{-1}(x)) = g^{-1}\left( \frac{x - 1}{2} \right) = \frac{\frac{x - 1}{2} + 3}{4} = \frac{x + 5}{8}$
3. Finally, apply $f^{-1}$ to the result: $f^{-1}\left( \frac{x + 5}{8} \right) = \frac{x + 5}{8} - 5 = \frac{x + 5 - 40}{8} = \frac{x - 35}{8}$ Since this doesn't match the given expression $\frac{x + 45}{8}$, Option A is incorrect.

B. $(f^{-1} \circ h^{-1} \circ g^{-1})(x) = \frac{x - 41}{8}$

We need to compute $f^{-1}(h^{-1}(g^{-1}(x)))$:

1. Start with $g^{-1}(x) = \frac{x + 3}{4}$.
2. Next, apply $h^{-1}$ to this: $h^{-1}(g^{-1}(x)) = h^{-1}\left( \frac{x + 3}{4} \right) = \frac{\frac{x + 3}{4} - 1}{2} = \frac{x - 1}{8}$
3. Finally, apply $f^{-1}$ to the result: $f^{-1}\left( \frac{x - 1}{8} \right) = \frac{x - 1}{8} - 5 = \frac{x - 1 - 40}{8} = \frac{x - 41}{8}$ This matches the given expression, so Option B is correct.

C. ( (g^{-1} \circ f^{-1} \circ h