Examine whether `x((a^(x)+1)/(a^(x)-1))` is an odd or even function.

Answer

To determine whether the function \( f(x) = x \cdot \frac{a^x + 1}{a^x - 1} \) is odd or even, we will follow these steps:

### Step 1: Define the function
Let’s define the function:
\[ f(x) = x \cdot \frac{a^x + 1}{a^x - 1} \]

### Step 2: Find \( f(-x) \)
Now, we need to calculate \( f(-x) \):
\[ f(-x) = -x \cdot \frac{a^{-x} + 1}{a^{-x} - 1} \]

### Step 3: Simplify \( f(-x) \)
We can rewrite \( a^{-x} \) as \( \frac{1}{a^x} \):
\[ f(-x) = -x \cdot \frac{\frac{1}{a^x} + 1}{\frac{1}{a^x} - 1} \]
Now, we can simplify the fraction:
\[ f(-x) = -x \cdot \frac{\frac{1 + a^x}{a^x}}{\frac{1 - a^x}{a^x}} = -x \cdot \frac{1 + a^x}{1 - a^x} \]

### Step 4: Factor out the negative sign
Now, we can factor out the negative sign:
\[ f(-x) = -x \cdot \frac{1 + a^x}{1 - a^x} \]

### Step 5: Compare \( f(-x) \) with \( -f(x) \)
Now, we need to check if \( f(-x) = -f(x) \):
\[ -f(x) = -\left(x \cdot \frac{a^x + 1}{a^x - 1}\right) = -x \cdot \frac{a^x + 1}{a^x - 1} \]
Notice that:
\[ f(-x) = -x \cdot \frac{1 + a^x}{1 - a^x} \]

### Step 6: Establish equality
Now, we can see that:
\[ f(-x) = -x \cdot \frac{1 + a^x}{1 - a^x} \quad \text{and} \quad -f(x) = -x \cdot \frac{a^x + 1}{a^x - 1} \]
Since \( \frac{1 + a^x}{1 - a^x} \) is not equal to \( \frac{a^x + 1}{a^x - 1} \), we need to check if they are equivalent.

### Step 7: Conclusion
Since \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), we conclude that:
\[ f(x) \text{ is neither odd nor even.} \]