If `f, g: R to R` be defined by `f(x)=2x+1` and `g(x) =x^(2)-2, AA x in R,` respectively. Then , find `gof` .

Explanation

To find \( g \circ f \) (read as "g of f"), we need to substitute the function \( f(x) \) into the function \( g(x) \). Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Identify the Functions**:
- We have \( f(x) = 2x + 1 \)
- We have \( g(x) = x^2 - 2 \)

2. **Find \( g(f(x)) \)**:
- We need to find \( g(f(x)) \), which means we will substitute \( f(x) \) into \( g(x) \).
- Therefore, \( g(f(x)) = g(2x + 1) \).

3. **Substitute \( f(x) \) into \( g(x) \)**:
- Now, substitute \( 2x + 1 \) into \( g(x) \):
\[ g(2x + 1) = (2x + 1)^2 - 2 \]

4. **Expand \( (2x + 1)^2 \)**:
- Use the identity \( (a + b)^2 = a^2 + 2ab + b^2 \):
\[ (2x + 1)^2 = (2x)^2 + 2 \cdot (2x) \cdot 1 + 1^2 = 4x^2 + 4x + 1 \]

5. **Subtract 2**:
- Now, we will subtract 2 from the expanded expression:
\[ g(2x + 1) = 4x^2 + 4x + 1 - 2 \]
\[ g(2x + 1) = 4x^2 + 4x - 1 \]

6. **Final Result**:
- Therefore, \( g \circ f = g(f(x)) = 4x^2 + 4x - 1 \).

### Summary:
The composition of the functions \( g \) and \( f \) is given by:
\[ g \circ f = 4x^2 + 4x - 1 \]