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

Options

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

Answer

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

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \).

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

Now, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}\left(x - \frac{1}{x}\right) = 1 + \frac{1}{x^2} \]

### Step 2: Set the derivative less than zero
To find where the function is decreasing, we need to set the derivative less than zero:

\[ f'(x) < 0 \]

This gives us:

\[ 1 + \frac{1}{x^2} < 0 \]

### Step 3: Analyze the inequality
The expression \( 1 + \frac{1}{x^2} \) is always positive for all \( x \neq 0 \) because \( \frac{1}{x^2} \) is always positive for any non-zero \( x \). Therefore, there are no values of \( x \) for which \( f'(x) < 0 \).

### Step 4: Conclusion
Since \( f'(x) \) is never less than zero for \( x \neq 0 \), the function \( f(x) \) does not decrease for any real values of \( x \).

Thus, the function \( f(x) \) does not decrease in any interval of its domain.