`f (x)=(x^(2)+1)/(x^(2)-1)` decreases in

Options

A. `(- infty, -1) uu (1, infty)`
B. `(0, 1) uu (1, infty)`
C. `R-{-1,1}`
D. none of these

Answer

To determine where the function \( f(x) = \frac{x^2 + 1}{x^2 - 1} \) is decreasing, we need to follow these steps:

### Step 1: Find the derivative of the function
We will use the quotient rule for differentiation, which states that if \( f(x) = \frac{u}{v} \), then \( f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \).

Let:
- \( u = x^2 + 1 \) and \( v = x^2 - 1 \)

Now, we compute the derivatives:
- \( u' = 2x \)
- \( v' = 2x \)

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

### Step 2: Simplify the derivative
Now, we simplify the numerator:
\[ f'(x) = \frac{(2x)(x^2 - 1) - (2x)(x^2 + 1)}{(x^2 - 1)^2} \]
\[ = \frac{2x(x^2 - 1 - x^2 - 1)}{(x^2 - 1)^2} \]
\[ = \frac{2x(-2)}{(x^2 - 1)^2} \]
\[ = \frac{-4x}{(x^2 - 1)^2} \]

### Step 3: Determine where the derivative is less than zero
To find where the function is decreasing, we need to solve:
\[ f'(x) < 0 \]
This gives us:
\[ \frac{-4x}{(x^2 - 1)^2} < 0 \]

Since the denominator \( (x^2 - 1)^2 \) is always positive (except where it is undefined), we can focus on the numerator:
\[ -4x < 0 \]
This implies:
\[ x > 0 \]

### Step 4: Consider the points where the function is undefined
The function \( f(x) \) is undefined when \( x^2 - 1 = 0 \), which gives us:
\[ x = \pm 1 \]
Thus, \( x \) cannot be \( 1 \) or \( -1 \).

### Step 5: Combine the intervals
From the analysis, we have:
- The function is decreasing for \( x > 0 \).
- We must exclude \( x = 1 \) from our interval.

Thus, the function \( f(x) \) decreases in the intervals:
\[ (0, 1) \cup (1, \infty) \]

### Final Answer
The function \( f(x) = \frac{x^2 + 1}{x^2 - 1} \) decreases in the intervals \( (0, 1) \) and \( (1, \infty) \).

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