The function `f(x) = x^(2) + gamma / x ` has a

Options

A. minimum at x = 2 if `gamma = 16`
B. maximum at x = 2 if `gamma = 16`
C. maximum for no real value of `gamma`
D. point of inflection at `x= 1 if gamma = -1`

Answer

To solve the problem, we need to analyze the function \( f(x) = x^2 + \frac{\gamma}{x} \) to determine its critical points, maxima, minima, and points of inflection.

### Step 1: Find the first derivative \( f'(x) \)

The first step is to differentiate the function with respect to \( x \).

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

Calculating the derivatives:

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

### Step 2: Set the first derivative to zero to find critical points

Next, we set the first derivative equal to zero to find the critical points.

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

Rearranging gives:

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

Multiplying both sides by \( x^2 \) yields:

\[ 2x^3 = \gamma \]

Thus, we can express \( x^3 \) as:

\[ x^3 = \frac{\gamma}{2} \tag{1} \]

### Step 3: Find the second derivative \( f''(x) \)

Now, we need to find the second derivative to determine the nature of the critical points.

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

Calculating the derivatives:

\[ f''(x) = 2 + \frac{2\gamma}{x^3} \]

### Step 4: Analyze the second derivative

To determine whether the critical point is a maximum or minimum, we substitute \( x^3 = \frac{\gamma}{2} \) into the second derivative.

\[ f''(x) = 2 + \frac{2\gamma}{\frac{\gamma}{2}} = 2 + 4 = 6 \]

Since \( f''(x) = 6 > 0 \), this indicates that the function has a local minimum at the critical point.

### Step 5: Determine the value of \( x \) when \( \gamma = 16 \)

To find the specific value of \( x \) when \( \gamma = 16 \):

From equation (1):

\[ x^3 = \frac{16}{2} = 8 \]

Taking the cube root gives:

\[ x = 2 \]

### Step 6: Determine the point of inflection

To find the point of inflection, we set the second derivative to zero:

\[ 2 + \frac{2\gamma}{x^3} = 0 \]

Rearranging gives:

\[ \frac{2\gamma}{x^3} = -2 \]

This leads to:

\[ \gamma = -\frac{x^3}{1} \]

Substituting \( \gamma = -1 \):

\[ x^3 = 1 \implies x = 1 \]

### Conclusion

1. The function has a local minimum at \( x = 2 \) when \( \gamma = 16 \).
2. The function has no maximum for any real value of \( \gamma \).
3. The point of inflection occurs at \( x = 1 \) when \( \gamma = -1 \).

### Final Answer

- Minimum at \( x = 2 \) when \( \gamma = 16 \) (Option A).
- No maximum for any real value of \( \gamma \) (Option C).
- Point of inflection at \( x = 1 \) when \( \gamma = -1 \) (Option D).