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

Options

A. x = 1
B. x = 2
C. x = - 2
D. x = - 1

Answer

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

### Step 1: Differentiate the function
We need to find the first derivative of the function \( f(x) \).

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

Calculating the derivatives:
- The derivative of \( \frac{x}{2} \) is \( \frac{1}{2} \).
- The derivative of \( \frac{2}{x} \) is \( -\frac{2}{x^2} \).

Thus, we have:

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

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

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

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

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

Cross-multiplying yields:

\[ x^2 = 4 \]

Taking the square root gives:

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

### Step 4: Determine the nature of the critical points
To determine whether these critical points are local minima or maxima, we need to find the second derivative of the function.

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

Calculating the second derivative:

\[ f''(x) = \frac{d}{dx}\left(\frac{1}{2} - \frac{2}{x^2}\right) = 0 + \frac{4}{x^3} \]

### Step 5: Evaluate the second derivative at the critical points
Now we evaluate \( f''(x) \) at \( x = 2 \) and \( x = -2 \):

1. For \( x = 2 \):

\[ f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} > 0 \]

2. For \( x = -2 \):

\[ f''(-2) = \frac{4}{(-2)^3} = \frac{4}{-8} = -\frac{1}{2} < 0 \]

### Conclusion
Since \( f''(2) > 0 \), there is a local minimum at \( x = 2 \).

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

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