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 step by step, we will analyze the function \( f(x) = x^2 + \frac{\gamma}{x} \) to determine its maxima, minima, and points of inflection.

### Step 1: Find the first derivative
We start by differentiating the function \( f(x) \).

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

Using the power rule and the quotient rule, we get:

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

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

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

Rearranging gives:

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

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

\[ 2x^3 = \gamma \]

Thus, we find:

\[ x^3 = \frac{\gamma}{2} \quad \Rightarrow \quad x = \sqrt[3]{\frac{\gamma}{2}} \]

### Step 3: Find the second derivative
Next, we find the second derivative to determine the nature of the critical points:

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

### Step 4: Analyze the second derivative
Now, we evaluate the second derivative at the critical point \( x = \sqrt[3]{\frac{\gamma}{2}} \):

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

Since \( \left(\sqrt[3]{\frac{\gamma}{2}}\right)^3 = \frac{\gamma}{2} \), we have:

\[ f''\left(\sqrt[3]{\frac{\gamma}{2}}\right) = 2 + \frac{2\gamma}{\frac{\gamma}{2}} = 2 + 4 = 6 \]

Since \( f''\left(\sqrt[3]{\frac{\gamma}{2}}\right) > 0 \), the critical point is a minimum.

### Step 5: Find the minimum value
Now, we substitute \( x = \sqrt[3]{\frac{\gamma}{2}} \) back into the original function to find the minimum value:

\[ f\left(\sqrt[3]{\frac{\gamma}{2}}\right) = \left(\sqrt[3]{\frac{\gamma}{2}}\right)^2 + \frac{\gamma}{\sqrt[3]{\frac{\gamma}{2}}} \]

Calculating this gives:

\[ = \frac{\gamma^{2/3}}{2^{2/3}} + \frac{\gamma \cdot 2^{1/3}}{\gamma^{1/3}} = \frac{\gamma^{2/3}}{2^{2/3}} + 2^{1/3} \gamma^{2/3} = \gamma^{2/3}\left(\frac{1}{2^{2/3}} + 2^{1/3}\right) \]

### Step 6: Determine points of inflection
To find points of inflection, we set the first derivative equal to zero:

\[ f'(x) = 2x - \frac{\gamma}{x^2} = 0 \Rightarrow 2x^3 = \gamma \Rightarrow x^3 = \frac{\gamma}{2} \]

### Conclusion
1. The function has a minimum at \( x = \sqrt[3]{\frac{\gamma}{2}} \).
2. The minimum value is \( f\left(\sqrt[3]{\frac{\gamma}{2}}\right) = \gamma^{2/3}\left(\frac{1}{2^{2/3}} + 2^{1/3}\right) \).
3. There are no maximum points for real values of \( x \).
4. The point of inflection occurs at \( x = \sqrt[3]{\frac{\gamma}{2}} \).