f(λx, λy) = (1/(ln(2λx + λy + 3) + 1)) - 2

The given function is:

\[
f(\lambda x, \lambda y) = \frac{1}{\ln(2\lambda x + \lambda y + 3) + 1} - 2
\]

### Let's break down the function step by step:

1. **Input:**
   The function takes two variables, \(\lambda x\) and \(\lambda y\), which are likely to be scaled forms of \(x\) and \(y\) by a parameter \(\lambda\).

2. **Inside the logarithmic function:**
   The expression inside the natural logarithm (\(\ln\)) is \(2\lambda x + \lambda y + 3\). This must be positive for the logarithm to be valid:
   \[
   2\lambda x + \lambda y + 3 > 0
   \]
   This implies that the function is only defined for values of \(x\) and \(y\) such that the argument of the logarithm is positive.

3. **Logarithmic term:**
   \[
   \ln(2\lambda x + \lambda y + 3)
   \]
   The natural logarithm increases as its input increases. Hence, larger values of \(2\lambda x + \lambda y + 3\) will result in larger values of the logarithmic term.

4. **Denominator:**
   The denominator is the logarithmic expression plus 1:
   \[
   \ln(2\lambda x + \lambda y + 3) + 1
   \]
   This ensures that the denominator is at least 1 (since the logarithmic term is always greater than or equal to 0).

5. **Main fraction:**
   The overall fraction becomes:
   \[
   \frac{1}{\ln(2\lambda x + \lambda y + 3) + 1}
   \]
   This fraction decreases as the logarithmic term increases. Therefore, as the input \(\lambda x\) or \(\lambda y\) grows, the function decreases.

6. **Final result:**
   Subtracting 2 from the fraction:
   \[
   f(\lambda x, \lambda y) = \frac{1}{\ln(2\lambda x + \lambda y + 3) + 1} - 2
   \]
   This final subtraction lowers the entire function value by 2 units.

### Behavior of the function:

- For large values of \(\lambda x\) and \(\lambda y\), the logarithmic term becomes large, making the fraction approach 0. Thus, the function approaches \(-2\).
- For smaller values of \(\lambda x\) and \(\lambda y\), the fraction is larger, meaning the function is greater than \(-2\).
- As \(2\lambda x + \lambda y + 3\) approaches 0 from above, the logarithmic term becomes very small, and the function value becomes larger.

Would you like further details, or do you have specific questions about the function's behavior or applications?

Here are some related questions to consider:

1. What happens to the function as \(\lambda x \to \infty\) or \(\lambda y \to \infty\)?
2. How can you solve for specific values of \(\lambda x\) and \(\lambda y\) where the function equals zero?
3. What are the critical points of this function in terms of \(x\) and \(y\)?
4. How does the function change as \(\lambda\) changes (e.g., \(\lambda = 1\), \(\lambda = 2\))?
5. What is the domain of the function in terms of \(x\) and \(y\)?

**Tip:** Always check the domain when working with logarithmic functions, as they are only defined for positive arguments.

Explore how to analyze and solve the logarithmic function f(λx, λy) = (1/(ln(2λx + λy + 3) + 1)) - 2. This problem covers key concepts in algebra and logarithmic functions, with a focus on understanding the behavior of the function as inputs scale. Suitable for high school students in Grades 10-12.

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