https://r2.erweima.ai/mathbot/image/1726392874012n982egz3.jpg

The problem presented in the image is as follows:

We are given that:

\[
f\left(\frac{x - 3}{x + 1}\right) = 2x
\]

The task is to determine \( f(x) \).

### Step-by-Step Solution:

1. **Substitute a new variable for the argument inside \( f \)**:
   Let 
   \[
   y = \frac{x - 3}{x + 1}
   \]
   We want to express \( x \) in terms of \( y \).

2. **Solve for \( x \) in terms of \( y \)**:
   \[
   y = \frac{x - 3}{x + 1}
   \]
   Multiply both sides by \( x + 1 \):
   \[
   y(x + 1) = x - 3
   \]
   Expand and rearrange:
   \[
   yx + y = x - 3
   \]
   Bring all \( x \)-terms to one side:
   \[
   yx - x = -3 - y
   \]
   Factor out \( x \):
   \[
   x(y - 1) = -3 - y
   \]
   Solve for \( x \):
   \[
   x = \frac{-3 - y}{y - 1}
   \]

3. **Substitute this \( x \) into the given expression \( f\left(\frac{x - 3}{x + 1}\right) = 2x \)**:
   Since we let \( y = \frac{x - 3}{x + 1} \), and from the given problem, \( f(y) = 2x \), we now substitute \( x = \frac{-3 - y}{y - 1} \):
   \[
   f(y) = 2\left(\frac{-3 - y}{y - 1}\right)
   \]

Thus, the function \( f(x) \) is:

\[
f(x) = 2 \cdot \frac{-3 - x}{x - 1}
\]

This is the required solution for \( f(x) \).

---

Would you like more details or explanations on any step?

Here are 5 related questions for further exploration:
1. How can we check if a function is injective or surjective based on its algebraic form?
2. What are some real-world applications of functional transformations like the one seen here?
3. How can we interpret the graph of \( f(x) = 2 \cdot \frac{-3 - x}{x - 1} \)?
4. What are some other methods to solve for a function given a transformation?
5. How would the solution change if the form of the equation were different, say \( f\left(\frac{x + a}{x + b}\right) = 2x \)?

**Tip**: Always verify your result by substituting it back into the original equation to check for consistency.

Learn how to determine the function f(x) from the equation f((x-3)/(x+1)) = 2x through a step-by-step algebraic solution. This problem involves concepts like functions, substitution, and algebraic manipulation, suitable for students in Grades 10-12.

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