The function `f(x) = x-1/x, x in R, x ne 0` is increasing

Answer

To determine whether the function \( f(x) = x - \frac{1}{x} \) is increasing for \( x \in \mathbb{R} \) and \( x \neq 0 \), we need to find the derivative of the function and analyze its sign.

### Step-by-step Solution:

1. **Define the function**:
\[ f(x) = x - \frac{1}{x} \]

2. **Find the derivative**:
To determine if the function is increasing, we need to compute the derivative \( f'(x) \).
\[ f'(x) = \frac{d}{dx}\left(x - \frac{1}{x}\right) \]
Using the power rule and the derivative of \( \frac{1}{x} \):
\[ f'(x) = 1 - \left(-\frac{1}{x^2}\right) = 1 + \frac{1}{x^2} \]

3. **Analyze the derivative**:
We need to check if \( f'(x) > 0 \) for all \( x \neq 0 \).
\[ f'(x) = 1 + \frac{1}{x^2} \]
Since \( x^2 > 0 \) for all \( x \neq 0 \), it follows that \( \frac{1}{x^2} > 0 \). Therefore:
\[ f'(x) = 1 + \frac{1}{x^2} > 1 > 0 \]

4. **Conclusion**:
Since \( f'(x) > 0 \) for all \( x \neq 0 \), the function \( f(x) \) is increasing for all \( x \in \mathbb{R} \) where \( x \neq 0 \).

Thus, the statement that the function \( f(x) = x - \frac{1}{x} \) is increasing is **true**.