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We are given three functions:

- \( f(x) = 2x + 5 \)
- \( g(x) = x^2 \)
- \( h(x) = \frac{x}{x - 4} \)

We are tasked to find \( (f \circ g \circ h)(6) \), which means we first need to compute \( h(6) \), then apply \( g \) to that result, and finally apply \( f \).

### Step 1: Calculate \( h(6) \)
Using the function \( h(x) = \frac{x}{x - 4} \), we substitute \( x = 6 \):

\[
h(6) = \frac{6}{6 - 4} = \frac{6}{2} = 3
\]

### Step 2: Calculate \( g(h(6)) = g(3) \)
Next, apply the function \( g(x) = x^2 \) to the result of \( h(6) = 3 \):

\[
g(3) = 3^2 = 9
\]

### Step 3: Calculate \( f(g(h(6))) = f(9) \)
Finally, apply the function \( f(x) = 2x + 5 \) to the result of \( g(3) = 9 \):

\[
f(9) = 2(9) + 5 = 18 + 5 = 23
\]

### Final Answer:
\[
(f \circ g \circ h)(6) = 23
\]

Would you like any further explanation on these steps?

Here are five related questions to expand this topic:
1. How would the result change if we calculated \( (g \circ h \circ f)(6) \)?
2. What happens if the value of \( x \) approaches 4 in the function \( h(x) \)?
3. Can you calculate \( (f \circ h \circ g)(2) \) using the same functions?
4. What are the domains of the functions \( f(x) \), \( g(x) \), and \( h(x) \)?
5. How would the function compositions differ if \( h(x) \) were defined differently?

**Tip:** When performing function compositions, work step-by-step from the innermost function outward to avoid mistakes.

Given the functions f(x) = 2x + 5, g(x) = x^2, and h(x) = x / (x - 4), find (f ∘ g ∘ h)(6).

Learn how to find the composition of functions f(x) = 2x + 5, g(x) = x^2, and h(x) = x / (x - 4) evaluated at x = 6. This problem covers important concepts in algebra and function composition, suitable for students in Grades 9-12.

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