consider the function `f(x)=(x^(2))/(x^(2)-1)`
The interval in which f is increasing is

Options

A. (-1,1)
B. `(-oo,-1)cup(-1,0)`
C. `(-oo,-oo)-{-1,1}`
D. `(0,1)cup(1,oo)`

Answer

To determine the interval in which the function \( f(x) = \frac{x^2}{x^2 - 1} \) is increasing, we need to follow these steps:

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

We will use the quotient rule for differentiation, which states that if \( f(x) = \frac{u}{v} \), then:

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

Here, let \( u = x^2 \) and \( v = x^2 - 1 \).

- The derivative of \( u \) is \( u' = 2x \).
- The derivative of \( v \) is \( v' = 2x \).

Now applying the quotient rule:

\[ f'(x) = \frac{(2x)(x^2 - 1) - (x^2)(2x)}{(x^2 - 1)^2} \]

### Step 2: Simplify the derivative

Now we simplify the numerator:

\[ f'(x) = \frac{2x(x^2 - 1) - 2x(x^2)}{(x^2 - 1)^2} \]

This simplifies to:

\[ f'(x) = \frac{2x^3 - 2x - 2x^3}{(x^2 - 1)^2} \]

The \( 2x^3 \) terms cancel out:

\[ f'(x) = \frac{-2x}{(x^2 - 1)^2} \]

### Step 3: Determine where \( f'(x) > 0 \)

For \( f(x) \) to be increasing, we need \( f'(x) > 0 \):

\[ \frac{-2x}{(x^2 - 1)^2} > 0 \]

Since \( (x^2 - 1)^2 \) is always positive (as it is a square), the sign of \( f'(x) \) depends solely on the numerator \( -2x \).

Thus, we require:

\[ -2x > 0 \implies x < 0 \]

### Step 4: Identify the intervals

The function \( f(x) \) is undefined at \( x = 1 \) and \( x = -1 \) (where the denominator becomes zero). Therefore, we need to consider the intervals around these points:

1. \( (-\infty, -1) \)
2. \( (-1, 0) \)
3. \( (0, 1) \)
4. \( (1, \infty) \)

From our analysis, \( f'(x) > 0 \) when \( x < 0 \). However, we also need to exclude the point \( x = -1 \) where the function is undefined.

Thus, the intervals where \( f(x) \) is increasing are:

\[ (-\infty, -1) \cup (-1, 0) \]

### Final Answer

The function \( f(x) \) is increasing in the intervals \( (-\infty, -1) \) and \( (-1, 0) \).

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