f(x) = 2x+2
g(x) = x² - 2
(f o g)(2) =...

To solve \((f \circ g)(2)\), which means applying \(f(g(2))\), we follow these steps:

### Step 1: Find \(g(2)\)
Using the function \(g(x) = x^2 - 2\):
\[
g(2) = 2^2 - 2 = 4 - 2 = 2
\]

### Step 2: Apply \(f\) to \(g(2)\)
Now, we apply \(f(x) = 2x + 2\) to the result \(g(2) = 2\):
\[
f(2) = 2(2) + 2 = 4 + 2 = 6
\]

Thus, \((f \circ g)(2) = 6\).

Would you like more details on the composition of functions or any other clarifications?

Here are five related questions for further exploration:
1. What is \((g \circ f)(2)\)?
2. How do you find the domain of \((f \circ g)(x)\)?
3. What is \((f \circ g)(x)\) in general form?
4. How does composition of functions affect their range?
5. What is the inverse of the function \(f(x) = 2x + 2\)?

**Tip**: Always check the domain and range of the individual functions before performing compositions to avoid undefined results.

f(x) = 2x + 2
g(x) = x² - 2
(f o g)(2) =...

Learn how to solve (f o g)(2) using function composition, where f(x) = 2x + 2 and g(x) = x² - 2. This problem demonstrates key algebraic concepts and is suitable for students in Grades 8-10.

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