Find the interval in which the function `f(x)=x/2+2/x, x < 0` is decreasing.

Answer

To find the interval in which the function \( f(x) = \frac{x}{2} + \frac{2}{x} \) is decreasing for \( x < 0 \), we will follow these steps:

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

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

Using the power rule and the quotient rule, we differentiate each term:

- The derivative of \( \frac{x}{2} \) is \( \frac{1}{2} \).
- The derivative of \( \frac{2}{x} \) can be rewritten as \( 2x^{-1} \), and its derivative is \( -2x^{-2} = -\frac{2}{x^2} \).

Thus, we have:

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

### Step 2: Set the derivative less than zero
To find where the function is decreasing, we set the derivative \( f'(x) \) less than zero:

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

### Step 3: Solve the inequality
Rearranging the inequality gives:

\[ \frac{2}{x^2} > \frac{1}{2} \]

Multiplying both sides by \( x^2 \) (noting that \( x^2 > 0 \) since \( x < 0 \)):

\[ 2 > \frac{1}{2} x^2 \]

Multiplying both sides by 2:

\[ 4 > x^2 \]

This can be rewritten as:

\[ x^2 < 4 \]

### Step 4: Find the values of \( x \)
Taking the square root of both sides gives:

\[ -\sqrt{4} < x < \sqrt{4} \]

This simplifies to:

\[ -2 < x < 2 \]

### Step 5: Consider the restriction \( x < 0 \)
Since we are only interested in the interval where \( x < 0 \), we take the part of the interval that satisfies this condition:

\[ -2 < x < 0 \]

### Conclusion
Thus, the function \( f(x) \) is decreasing in the interval:

\[ (-2, 0) \]