If `f(x) =2x +1 and g (x) (x-1)/(2)` for all real x, then `(fog) ^(-1)( ((1)/(x))` is equal to

Options

A. x
B. `1/x`
C. `-x`
D. `-1/x`

Answer

To solve the problem, we need to find the inverse of the composition of two functions \( f \) and \( g \), specifically \( (f \circ g)^{-1}\left(\frac{1}{x}\right) \).

### Step-by-Step Solution:

1. **Define the Functions:**
- Given \( f(x) = 2x + 1 \)
- Given \( g(x) = \frac{x - 1}{2} \)

2. **Find the Composition \( f(g(x)) \):**
- We need to substitute \( g(x) \) into \( f(x) \).
- So, \( f(g(x)) = f\left(\frac{x - 1}{2}\right) \).
- Now, replace \( x \) in \( f(x) \):
\[ f(g(x)) = 2\left(\frac{x - 1}{2}\right) + 1 \]
- Simplifying this:
\[ = (x - 1) + 1 = x \]

3. **Find the Inverse of \( f(g(x)) \):**
- We have found that \( f(g(x)) = x \).
- To find the inverse, we set \( y = f(g(x)) \) and solve for \( x \):
\[ y = x \implies x = y \]
- Therefore, \( (f \circ g)^{-1}(y) = y \).

4. **Evaluate \( (f \circ g)^{-1}\left(\frac{1}{x}\right) \):**
- Now we need to find \( (f \circ g)^{-1}\left(\frac{1}{x}\right) \):
\[ (f \circ g)^{-1}\left(\frac{1}{x}\right) = \frac{1}{x} \]

### Final Answer:
Thus, the final result is:
\[ (f \circ g)^{-1}\left(\frac{1}{x}\right) = \frac{1}{x} \]