Let `f (x) = (alpha x)/( x+1) , x ne -1` . Then, for what value of `alpha` is `f{f (x)} = x?`

Options

A. `sqrt(2)`
B. `- sqrt(2)`
C. `1`
D. `-1`

Answer

To solve the problem, we need to find the value of \( \alpha \) such that \( f(f(x)) = x \), where \( f(x) = \frac{\alpha x}{x + 1} \).

### Step-by-step Solution:

1. **Define the function**:
\[ f(x) = \frac{\alpha x}{x + 1} \]

2. **Find \( f(f(x)) \)**:
We need to substitute \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{\alpha x}{x + 1}\right) \]
Substitute \( \frac{\alpha x}{x + 1} \) into \( f(x) \):
\[ f\left(\frac{\alpha x}{x + 1}\right) = \frac{\alpha \left(\frac{\alpha x}{x + 1}\right)}{\frac{\alpha x}{x + 1} + 1} \]

3. **Simplify the denominator**:
The denominator can be simplified as follows:
\[ \frac{\alpha x}{x + 1} + 1 = \frac{\alpha x + (x + 1)}{x + 1} = \frac{(\alpha + 1)x + 1}{x + 1} \]

4. **Substituting back**:
Now, substituting this back into the function:
\[ f(f(x)) = \frac{\alpha \left(\frac{\alpha x}{x + 1}\right)}{\frac{(\alpha + 1)x + 1}{x + 1}} = \frac{\alpha^2 x}{(\alpha + 1)x + 1} \]

5. **Set \( f(f(x)) = x \)**:
We set the expression equal to \( x \):
\[ \frac{\alpha^2 x}{(\alpha + 1)x + 1} = x \]

6. **Cross-multiply**:
Cross-multiplying gives:
\[ \alpha^2 x = x \cdot ((\alpha + 1)x + 1) \]
Expanding the right-hand side:
\[ \alpha^2 x = (\alpha + 1)x^2 + x \]

7. **Rearranging the equation**:
Rearranging gives:
\[ (\alpha + 1)x^2 + (1 - \alpha^2)x = 0 \]

8. **Factoring out \( x \)**:
Factoring out \( x \):
\[ x \left((\alpha + 1)x + (1 - \alpha^2)\right) = 0 \]
For this equation to hold for all \( x \), the coefficient of \( x \) must equal zero:
\[ \alpha + 1 = 0 \quad \text{and} \quad 1 - \alpha^2 = 0 \]

9. **Solving the equations**:
From \( \alpha + 1 = 0 \):
\[ \alpha = -1 \]
From \( 1 - \alpha^2 = 0 \):
\[ \alpha^2 = 1 \implies \alpha = 1 \text{ or } \alpha = -1 \]

10. **Conclusion**:
The values of \( \alpha \) that satisfy \( f(f(x)) = x \) are \( \alpha = 1 \) or \( \alpha = -1 \).

### Final Answer:
The values of \( \alpha \) for which \( f(f(x)) = x \) are \( \alpha = 1 \) and \( \alpha = -1 \).

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