If `f: R to R and g: R to R ` be two functions defined as f(x)=2x+1 and `g(x)=x^(2)-2` respectively , then find (gof) (x) and (fog) (x) and show that (fog) (x) `ne` (gof) (x).

Answer

To solve the problem, we need to find the compositions of the functions \( f \) and \( g \) defined as follows:

- \( f(x) = 2x + 1 \)
- \( g(x) = x^2 - 2 \)

We need to find \( (g \circ f)(x) \) and \( (f \circ g)(x) \), and then show that \( (f \circ g)(x) \neq (g \circ f)(x) \).

### Step 1: Find \( (g \circ f)(x) \)

The composition \( (g \circ f)(x) \) means we need to apply \( f \) first and then apply \( g \) to the result of \( f \).

1. Start with \( f(x) \):
\[ f(x) = 2x + 1 \]

2. Now substitute \( f(x) \) into \( g \):
\[ g(f(x)) = g(2x + 1) \]

3. Using the definition of \( g(x) \):
\[ g(x) = x^2 - 2 \]
Therefore,
\[ g(2x + 1) = (2x + 1)^2 - 2 \]

4. Expand \( (2x + 1)^2 \):
\[ (2x + 1)^2 = 4x^2 + 4x + 1 \]

5. Now substitute this back into the equation:
\[ g(2x + 1) = 4x^2 + 4x + 1 - 2 = 4x^2 + 4x - 1 \]

Thus, we have:
\[ (g \circ f)(x) = 4x^2 + 4x - 1 \]

### Step 2: Find \( (f \circ g)(x) \)

Now we find \( (f \circ g)(x) \), which means we apply \( g \) first and then apply \( f \) to the result of \( g \).

1. Start with \( g(x) \):
\[ g(x) = x^2 - 2 \]

2. Now substitute \( g(x) \) into \( f \):
\[ f(g(x)) = f(x^2 - 2) \]

3. Using the definition of \( f(x) \):
\[ f(x) = 2x + 1 \]
Therefore,
\[ f(x^2 - 2) = 2(x^2 - 2) + 1 \]

4. Simplify the expression:
\[ f(x^2 - 2) = 2x^2 - 4 + 1 = 2x^2 - 3 \]

Thus, we have:
\[ (f \circ g)(x) = 2x^2 - 3 \]

### Step 3: Compare \( (g \circ f)(x) \) and \( (f \circ g)(x) \)

Now we compare the two results:
- \( (g \circ f)(x) = 4x^2 + 4x - 1 \)
- \( (f \circ g)(x) = 2x^2 - 3 \)

To show that they are not equal, we can set them equal to each other and check if there are any solutions:
\[ 4x^2 + 4x - 1 \neq 2x^2 - 3 \]

Rearranging gives:
\[ 4x^2 + 4x - 1 - 2x^2 + 3 \neq 0 \]
\[ 2x^2 + 4x + 2 \neq 0 \]

This is a quadratic equation. To determine if it has any real solutions, we can check the discriminant:
\[ D = b^2 - 4ac = 4^2 - 4 \cdot 2 \cdot 2 = 16 - 16 = 0 \]

Since the discriminant is zero, it means the quadratic has a double root, which indicates that \( (g \circ f)(x) \) and \( (f \circ g)(x) \) are not equal for all \( x \).

### Conclusion

Thus, we have shown that:
\[ (f \circ g)(x) \neq (g \circ f)(x) \]