Find the interval in which `f(x)=log_eabs((x-1)/(x+1))-2/(x-1)` is increasing

Options

A. `(-oo,-1)uu[0,1)uu(1,oo)`
B. `(-1,0)uu[0,1)uu(1,oo)`
C. `(-1,-oo)`
D. `(-oo,-1)uu(-1,1)`

Answer

To find the interval in which the function \( f(x) = \log_e \left| \frac{x-1}{x+1} \right| - \frac{2}{x-1} \) is increasing, we need to follow these steps:

### Step 1: Find the derivative \( f'(x) \)

To determine where the function is increasing, we first need to compute the derivative \( f'(x) \).

1. The derivative of \( \log_e \left| \frac{x-1}{x+1} \right| \) can be computed using the chain rule and the quotient rule.
2. The derivative of \( -\frac{2}{x-1} \) is \( \frac{2}{(x-1)^2} \).

Using the quotient rule for the logarithmic part:
\[ f'(x) = \frac{1}{\frac{x-1}{x+1}} \cdot \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} + \frac{2}{(x-1)^2} \]

This simplifies to:
\[ f'(x) = \frac{(x+1) - (x-1)}{(x-1)(x+1)} + \frac{2}{(x-1)^2} \]
\[ = \frac{2}{(x-1)(x+1)} + \frac{2}{(x-1)^2} \]

### Step 2: Set the derivative greater than zero

To find where \( f(x) \) is increasing, we set \( f'(x) > 0 \):
\[ \frac{2}{(x-1)(x+1)} + \frac{2}{(x-1)^2} > 0 \]

### Step 3: Find a common denominator

The common denominator is \( (x-1)^2(x+1) \):
\[ \frac{2(x-1) + 2(x+1)}{(x-1)^2(x+1)} > 0 \]
\[ \frac{2x - 2 + 2x + 2}{(x-1)^2(x+1)} > 0 \]
\[ \frac{4x}{(x-1)^2(x+1)} > 0 \]

### Step 4: Analyze the sign of the expression

The expression \( \frac{4x}{(x-1)^2(x+1)} > 0 \) implies:
1. The numerator \( 4x > 0 \) gives \( x > 0 \).
2. The denominator \( (x-1)^2 > 0 \) is always positive except at \( x = 1 \).
3. The term \( (x+1) > 0 \) gives \( x > -1 \).

### Step 5: Combine the conditions

From the conditions:
- \( x > 0 \) (from the numerator)
- \( x \neq 1 \) (from the denominator)
- \( x > -1 \) (from the denominator)

The critical points to consider are \( x = 0 \) and \( x = 1 \).

### Step 6: Determine the intervals

The intervals based on the analysis are:
1. \( (-\infty, -1) \) (not valid since \( x > 0 \))
2. \( (0, 1) \) (valid)
3. \( (1, \infty) \) (valid)

Thus, the function is increasing in the intervals:
\[ (-\infty, -1) \cup (0, 1) \cup (1, \infty) \]

### Final Answer

The intervals in which \( f(x) \) is increasing are:
\[ (-\infty, -1) \cup (0, 1) \cup (1, \infty) \]