Show that the function `x- (4)/(x)` always increases as x increases.

Answer

To show that the function \( f(x) = x - \frac{4}{x} \) always increases as \( x \) increases, we will follow these steps:

### Step 1: Define the function
We start with the function:
\[ f(x) = x - \frac{4}{x} \]

### Step 2: Find the derivative of the function
To determine if the function is increasing, we need to find its derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}\left(x - \frac{4}{x}\right) \]
Using the derivative rules, we get:
\[ f'(x) = 1 - \frac{d}{dx}\left(\frac{4}{x}\right) \]
The derivative of \( \frac{4}{x} \) is \( -\frac{4}{x^2} \), so:
\[ f'(x) = 1 + \frac{4}{x^2} \]

### Step 3: Analyze the derivative
Next, we analyze \( f'(x) \):
\[ f'(x) = 1 + \frac{4}{x^2} \]
Since \( x^2 \) is always positive for all \( x \neq 0 \), \( \frac{4}{x^2} \) is also positive. Therefore, we can conclude that:
\[ f'(x) > 0 \quad \text{for all } x \neq 0 \]

### Step 4: Conclusion
Since the derivative \( f'(x) \) is positive for all \( x \neq 0 \), it indicates that the function \( f(x) \) is monotonically increasing in its domain. Thus, we can conclude that:
\[ \text{The function } f(x) = x - \frac{4}{x} \text{ always increases as } x \text{ increases.} \]