`f(x)= y+(1)/(y)`, where `y gt 0`. If y increases in value, then f (x)

Options

A. Fluctuates up and down in value
B. Decreases in values
C. Increases in value but at a much higher rate than y
D. Increases in value and tends towards y

Answer

To solve the problem, we need to analyze the function \( f(x) = y + \frac{1}{y} \) where \( y > 0 \). We want to determine how \( f(x) \) behaves as \( y \) increases.

### Step-by-Step Solution:

1. **Understand the Function**: The function given is \( f(x) = y + \frac{1}{y} \). Here, \( y \) is a variable that is greater than 0.

2. **Evaluate the Function for Different Values of \( y \)**:
- Let’s calculate \( f(x) \) for a few values of \( y \):
- For \( y = 0.5 \):
\[ f(0.5) = 0.5 + \frac{1}{0.5} = 0.5 + 2 = 2.5 \]
- For \( y = 1 \):
\[ f(1) = 1 + \frac{1}{1} = 1 + 1 = 2 \]
- For \( y = 2 \):
\[ f(2) = 2 + \frac{1}{2} = 2 + 0.5 = 2.5 \]
- For \( y = 3 \):
\[ f(3) = 3 + \frac{1}{3} = 3 + 0.333 \approx 3.333 \]

3. **Analyze the Results**:
- From the calculations, we can see that as \( y \) increases from \( 0.5 \) to \( 1 \), \( f(x) \) decreases from \( 2.5 \) to \( 2 \).
- As \( y \) continues to increase to \( 2 \) and \( 3 \), \( f(x) \) increases again to \( 2.5 \) and then to approximately \( 3.333 \).

4. **Conclusion**:
- The function \( f(x) \) does not consistently increase or decrease. Instead, it fluctuates based on the value of \( y \).
- Therefore, we can conclude that \( f(x) \) fluctuates as \( y \) increases.

### Final Answer:
As \( y \) increases, \( f(x) \) fluctuates.