The function `f(x) = (2)/x + x/2` has a local minimum at x =

Options

A. `-2`
B. 0
C. 1
D. 2

Answer

To find the local minimum of the function \( f(x) = \frac{2}{x} + \frac{x}{2} \), we will follow these steps:

### Step 1: Find the first derivative of the function
We start by differentiating the function with respect to \( x \).

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

Using the power rule and the derivative of \( \frac{1}{x} \):

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

### Step 2: Set the first derivative equal to zero
To find the critical points, we set the first derivative equal to zero:

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

### Step 3: Solve for \( x \)
Rearranging the equation gives:

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

Cross-multiplying results in:

\[ x^2 = 4 \]

Taking the square root of both sides, we find:

\[ x = 2 \quad \text{or} \quad x = -2 \]

### Step 4: Determine the nature of the critical point
Next, we need to determine whether \( x = 2 \) is a local minimum or maximum by finding the second derivative.

### Step 5: Find the second derivative
We differentiate the first derivative:

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

Calculating the derivative:

\[ f''(x) = \frac{4}{x^3} \]

### Step 6: Evaluate the second derivative at \( x = 2 \)
Now we substitute \( x = 2 \) into the second derivative:

\[ f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} \]

Since \( f''(2) > 0 \), this indicates that \( x = 2 \) is a point of local minimum.

### Conclusion
Thus, the function \( f(x) = \frac{2}{x} + \frac{x}{2} \) has a local minimum at \( x = 2 \).

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