If `g(x)` is a curve which is obtained by the reflection of `f(x) = (e^x - e^(-x)) / 2` then by the line `y = x` then

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{e^x - e^{-x}}{2} \) and its reflection across the line \( y = x \). The reflection of a function across the line \( y = x \) gives us its inverse function. Therefore, we need to find the inverse of \( f(x) \).

### Step 1: Set up the equation
Let \( y = f(x) \):
\[ y = \frac{e^x - e^{-x}}{2} \]

### Step 2: Solve for \( x \) in terms of \( y \)
Multiply both sides by 2:
\[ 2y = e^x - e^{-x} \]
Rearranging gives:
\[ e^x - 2y - e^{-x} = 0 \]
Now, multiply through by \( e^x \) to eliminate the negative exponent:
\[ e^{2x} - 2ye^x - 1 = 0 \]

### Step 3: Use the quadratic formula
This is a quadratic equation in terms of \( e^x \):
\[ e^{2x} - 2ye^x - 1 = 0 \]
Let \( u = e^x \). The equation becomes:
\[ u^2 - 2yu - 1 = 0 \]
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{2y \pm \sqrt{(2y)^2 + 4}}{2} \]
\[ u = y \pm \sqrt{y^2 + 1} \]

### Step 4: Solve for \( x \)
Since \( u = e^x \), we take the positive root (as \( e^x \) is always positive):
\[ e^x = y + \sqrt{y^2 + 1} \]
Taking the natural logarithm:
\[ x = \ln(y + \sqrt{y^2 + 1}) \]

### Step 5: Write the inverse function
Thus, the inverse function \( g(x) \) is:
\[ g(x) = \ln(x + \sqrt{x^2 + 1}) \]

### Step 6: Analyze the properties of \( g(x) \)
1. **Increasing or Decreasing**: To check if \( g(x) \) is always increasing, we can differentiate \( g(x) \):
\[ g'(x) = \frac{1}{x + \sqrt{x^2 + 1}} \cdot \left(1 + \frac{x}{\sqrt{x^2 + 1}}\right) \]
Since both terms in the product are positive, \( g'(x) > 0 \) for all \( x \). Thus, \( g(x) \) is always increasing.

2. **Extrema**: Since \( g(x) \) is always increasing, it has no local maxima or minima (no extrema).

### Conclusion
The correct option based on the analysis is that \( g(x) \) has no extremum and is always increasing.