`f(x)=((e^(2x)-1)/(e^(2x)+1))` is

Options

A. an increasing function
B. a decreasing function
C. an even function
D. None of these

Explanation

To determine the nature of the function \( f(x) = \frac{e^{2x} - 1}{e^{2x} + 1} \), we will check if it is an even function, an odd function, or if it is increasing or decreasing.

### Step 1: Check if the function is odd or even
1. **Calculate \( f(-x) \)**:
\[ f(-x) = \frac{e^{-2x} - 1}{e^{-2x} + 1} \]
We can rewrite \( e^{-2x} \) as \( \frac{1}{e^{2x}} \):
\[ f(-x) = \frac{\frac{1}{e^{2x}} - 1}{\frac{1}{e^{2x}} + 1} = \frac{1 - e^{2x}}{1 + e^{2x}} \]

2. **Rewrite \( f(-x) \)**:
\[ f(-x) = \frac{-(e^{2x} - 1)}{e^{2x} + 1} = -\frac{e^{2x} - 1}{e^{2x} + 1} = -f(x) \]
Since \( f(-x) = -f(x) \), the function is **odd**.

### Step 2: Find the derivative \( f'(x) \)
To determine if the function is increasing or decreasing, we need to find the derivative \( f'(x) \) using the quotient rule:
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
where \( u = e^{2x} - 1 \) and \( v = e^{2x} + 1 \).

1. **Calculate \( u' \) and \( v' \)**:
- \( u' = 2e^{2x} \)
- \( v' = 2e^{2x} \)

2. **Apply the quotient rule**:
\[ f'(x) = \frac{(2e^{2x})(e^{2x} + 1) - (e^{2x} - 1)(2e^{2x})}{(e^{2x} + 1)^2} \]

3. **Simplify**:
\[ f'(x) = \frac{2e^{2x}(e^{2x} + 1) - 2e^{2x}(e^{2x} - 1)}{(e^{2x} + 1)^2} \]
\[ = \frac{2e^{2x}(e^{2x} + 1 - e^{2x} + 1)}{(e^{2x} + 1)^2} \]
\[ = \frac{2e^{2x}(2)}{(e^{2x} + 1)^2} = \frac{4e^{2x}}{(e^{2x} + 1)^2} \]

### Step 3: Analyze the sign of \( f'(x) \)
Since \( e^{2x} > 0 \) for all \( x \), we have:
\[ f'(x) = \frac{4e^{2x}}{(e^{2x} + 1)^2} > 0 \]
This implies that \( f'(x) > 0 \) for all \( x \in \mathbb{R} \), meaning that the function is **increasing**.

### Conclusion
The function \( f(x) = \frac{e^{2x} - 1}{e^{2x} + 1} \) is an **odd function** and is **increasing** for all \( x \in \mathbb{R}**.