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.