The problem in the image asks to determine the reflection of the function $f(x) = \frac{x - 1}{4x + 2}, x \neq -\frac{1}{2}$, across two different lines:

1. Reflection over the line $y = x$.
2. Reflection over the line $y = -x$.

1. Reflection over the line $y = x$:

To reflect the function $f(x)$ over the line $y = x$, we swap $x$ and $y$, and solve for $y$.

Given: $y = \frac{x - 1}{4x + 2}$ We swap $x$ and $y$: $x = \frac{y - 1}{4y + 2}$ Now solve for $y$: $x(4y + 2) = y - 1$ $4xy + 2x = y - 1$ $4xy - y = -2x - 1$ $y(4x - 1) = -2x - 1$ $y = \frac{-2x - 1}{4x - 1}$ Thus, the equation of the reflection of $f(x)$ over the line $y = x$ is: $f^{-1}(x) = \frac{-2x - 1}{4x - 1}$

2. Reflection over the line $y = -x$:

To reflect $f(x)$ over the line $y = -x$, we follow a similar process, but now swap $x$ with $-y$.

Starting with: $y = \frac{x - 1}{4x + 2}$ Swap $x$ and $-y$: $x = \frac{-y - 1}{4(-y) + 2} = \frac{-y - 1}{-4y + 2}$ Solve for $y$: $x(-4y + 2) = -y - 1$ $-4xy + 2x = -y - 1$ $-4xy + y = -2x - 1$ $y(-4x + 1) = -2x - 1$ $y = \frac{-2x - 1}{-4x + 1} = \frac{2x + 1}{4x - 1}$ Thus, the reflection of $f(x)$ over the line $y = -x$ is: $f_{refl}(x) = \frac{2x + 1}{4x - 1}$

Would you like further clarification on any step? Also, here are some related questions you could explore:

1. How do you find the inverse of a general function?
2. What happens to the domain and range when a function is reflected over $y = x$?
3. How do reflections affect the asymptotes of rational functions?
4. What are the graphical implications of reflecting a function over $y = -x$?
5. How can we verify the correctness of a reflection algebraically?

Tip: Reflections over lines like $y = x$ or $y = -x$ often involve swapping variables and solving, making inverse function techniques particularly useful.