Let's analyze the mathematical problem in the uploaded image. The question states:

The function $f(x) = \frac{2 - 3x}{4x + 1}$ is given. If $f^{-1}$ is the inverse of the function $f(x)$, find $f^{-1}(x - 2)$.

Solution:

1. Find the inverse of $f(x)$: For $y = \frac{2 - 3x}{4x + 1}$, swap $x$ and $y$: $x = \frac{2 - 3y}{4y + 1}.$ Solve for $y$ (the inverse function): Multiply through by $4y + 1$: $x(4y + 1) = 2 - 3y.$ Expand: $4xy + x = 2 - 3y.$ Rearrange terms to isolate $y$: $4xy + 3y = 2 - x.$ Factor $y$ on the left: $y(4x + 3) = 2 - x.$ Solve for $y$: $y = \frac{2 - x}{4x + 3}.$ Thus, the inverse function is: $f^{-1}(x) = \frac{2 - x}{4x + 3}.$

2. Evaluate $f^{-1}(x - 2)$: Substitute $x - 2$ into $f^{-1}(x)$: $f^{-1}(x - 2) = \frac{2 - (x - 2)}{4(x - 2) + 3}.$ Simplify the numerator: $2 - (x - 2) = 2 - x + 2 = 4 - x.$ Simplify the denominator: $4(x - 2) + 3 = 4x - 8 + 3 = 4x - 5.$ Therefore: $f^{-1}(x - 2) = \frac{4 - x}{4x - 5}.$

Final Answer:

$f^{-1}(x - 2) = \frac{4 - x}{4x - 5}.$

Would you like a detailed explanation of any specific step? 😊

Related Questions:

1. How do you verify if two functions are inverses of each other?
2. What are the properties of inverse functions?
3. How do we find the domain and range of $f^{-1}(x)$?
4. What happens if $f(x)$ is not one-to-one? Can it still have an inverse?
5. How do transformations like $f(x - a)$ affect the inverse function?

Tip:

Always verify an inverse function by substituting it back into the original function: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.