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The problem asks to evaluate the following compositions of functions \( f(x) = 1 + x^2 \) and \( g(x) = \sqrt{x - 1} \):

a. \( f \circ g \)

b. \( g \circ f \)

c. \( (g \circ f)(2) \)

d. \( (f \circ f)(2) \)

Let's solve them step by step:

### a. \( f \circ g \)
This means \( (f \circ g)(x) = f(g(x)) \). Substituting \( g(x) = \sqrt{x - 1} \) into \( f(x) \):

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

Thus, \( (f \circ g)(x) = x \).

### b. \( g \circ f \)
This means \( (g \circ f)(x) = g(f(x)) \). Substituting \( f(x) = 1 + x^2 \) into \( g(x) \):

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

Thus, \( (g \circ f)(x) = |x| \).

### c. \( (g \circ f)(2) \)
Now, we compute \( (g \circ f)(2) \):

\[
f(2) = 1 + 2^2 = 1 + 4 = 5
\]
\[
g(5) = \sqrt{5 - 1} = \sqrt{4} = 2
\]

Thus, \( (g \circ f)(2) = 2 \).

### d. \( (f \circ f)(2) \)
Finally, we compute \( (f \circ f)(2) \):

\[
f(2) = 1 + 2^2 = 1 + 4 = 5
\]
\[
f(f(2)) = f(5) = 1 + 5^2 = 1 + 25 = 26
\]

Thus, \( (f \circ f)(2) = 26 \).

### Final answers:
a. \( (f \circ g)(x) = x \)

b. \( (g \circ f)(x) = |x| \)

c. \( (g \circ f)(2) = 2 \)

d. \( (f \circ f)(2) = 26 \)

Would you like further details on any part of the solution?

Here are some related questions:
1. What is the general process for function composition?
2. How does the domain of a composed function change?
3. What happens when you compose a function with its inverse?
4. Can you always compose two functions?
5. How do compositions apply in real-world problems?

**Tip:** When working with compositions, it's crucial to respect the domain constraints of both functions.

Let f(x) = 1 + x^2 and g(x) = sqrt(x - 1). Find the following: a. f ∘ g b. g ∘ f c. (g ∘ f)(2) d. (f ∘ f)(2)

Solve function composition problems with f(x) = 1 + x^2 and g(x) = sqrt(x - 1). This step-by-step solution explains how to compute f ∘ g, g ∘ f, and evaluate compositions at specific points. Suitable for Grades 10-12.

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