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

Options

A. an increasing function R
B. a decreasing function on R
C. an even function on R
D. none of these

Answer

To determine whether the function \( f(x) = \frac{e^{2x} - 1}{e^{2x} + 1} \) is increasing or decreasing, we will follow these steps:

### Step 1: Simplify the function
We start with the given function:
\[ f(x) = \frac{e^{2x} - 1}{e^{2x} + 1} \]
This function is already in a simplified form, so we can proceed to differentiate it.

### Step 2: Differentiate the function
We will use the quotient rule to differentiate \( f(x) \). The quotient rule states that if you have a function \( \frac{u}{v} \), then its derivative is given by:
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
Here, let \( u = e^{2x} - 1 \) and \( v = e^{2x} + 1 \).

Calculating the derivatives:
- \( u' = 2e^{2x} \)
- \( v' = 2e^{2x} \)

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

### Step 3: Simplify the derivative
Now we simplify the numerator:
\[ = \frac{2e^{2x}(e^{2x} + 1) - 2e^{2x}(e^{2x} - 1)}{(e^{2x} + 1)^2} \]
Distributing \( 2e^{2x} \):
\[ = \frac{2e^{4x} + 2e^{2x} - 2e^{4x} + 2e^{2x}}{(e^{2x} + 1)^2} \]
This simplifies to:
\[ = \frac{4e^{2x}}{(e^{2x} + 1)^2} \]

### Step 4: Analyze the sign of the derivative
Now we analyze \( f'(x) \):
\[ f'(x) = \frac{4e^{2x}}{(e^{2x} + 1)^2} \]
Since \( e^{2x} > 0 \) for all real \( x \), both the numerator \( 4e^{2x} \) and the denominator \( (e^{2x} + 1)^2 \) are always positive. Therefore, \( f'(x) > 0 \) for all real \( x \).

### Step 5: Conclusion
Since the derivative \( f'(x) > 0 \) for all \( x \), we conclude that the function \( f(x) \) is an increasing function for all real numbers.

### Final Answer
Thus, the function \( f(x) = \frac{e^{2x} - 1}{e^{2x} + 1} \) is an increasing function for all real numbers.

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