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We are asked to find the rule for \( f \circ g \), where \( f(x) = \sqrt{2x - 1} \) and \( g(x) = x^2 + 2 \). This composition of functions means we need to calculate \( f(g(x)) \).

Let's break it down:

### Step 1: Evaluate \( g(x) \)
We know that:
\[
g(x) = x^2 + 2
\]

### Step 2: Substitute \( g(x) \) into \( f(x) \)
The composition \( f \circ g \) means substituting \( g(x) \) into \( f(x) \). Since \( f(x) = \sqrt{2x - 1} \), we replace \( x \) in \( f(x) \) with \( g(x) \):
\[
f(g(x)) = \sqrt{2(g(x)) - 1}
\]
Substitute \( g(x) = x^2 + 2 \):
\[
f(g(x)) = \sqrt{2(x^2 + 2) - 1}
\]

### Step 3: Simplify the expression
Now, simplify the expression inside the square root:
\[
f(g(x)) = \sqrt{2x^2 + 4 - 1} = \sqrt{2x^2 + 3}
\]

Thus, the rule for \( f \circ g \) is:
\[
f \circ g(x) = \sqrt{2x^2 + 3}
\]

### Conclusion:
The correct answer is \( \boxed{\sqrt{2x^2 + 3}} \).

Do you have any questions or would you like further details on this solution?

Here are 5 follow-up questions to enhance your understanding:

1. What is the definition of a function composition, and how does it apply to \( f \circ g(x) \)?
2. How would the solution change if \( g(x) \) was a different function, like \( g(x) = x^2 - 1 \)?
3. What are the domain restrictions for the composite function \( f \circ g(x) \) in this case?
4. Can you derive the derivative of the composite function \( f \circ g(x) \)?
5. How would you check whether \( f \circ g(x) \) is injective or surjective?

**Tip:** When composing functions, carefully substitute the inner function and simplify step-by-step to avoid mistakes.

Find the rule for f ∘ g for f(x) = √(2x − 1) and g(x) = x² + 2.

This problem explores the composition of functions f(x) = √(2x − 1) and g(x) = x² + 2. Follow the steps to simplify the expression and find the composite function f ∘ g(x) = √(2x² + 3). This problem covers key algebraic concepts and is suitable for students in Grades 9-11.

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