Let `f(x) =(alphax)/(x+1)` Then the value of `alpha` for which `f(f(x) = x` is

Explanation

To solve the problem, we need to find the value of \( \alpha \) for which \( f(f(x)) = x \), 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:
\[ f(f(x)) = f\left(\frac{\alpha x}{x + 1}\right) \]
Now, replace \( x \) in the function with \( \frac{\alpha x}{x + 1} \):
\[ f\left(\frac{\alpha x}{x + 1}\right) = \frac{\alpha \left(\frac{\alpha x}{x + 1}\right)}{\frac{\alpha x}{x + 1} + 1} \]

3. **Simplify the denominator**:
The denominator becomes:
\[ \frac{\alpha x}{x + 1} + 1 = \frac{\alpha x + (x + 1)}{x + 1} = \frac{(\alpha + 1)x + 1}{x + 1} \]

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

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

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

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

8. **Factoring out \( x \)**:
This can be factored as:
\[ x \left((\alpha + 1)x + (1 - \alpha^2)\right) = 0 \]
For this equation to hold for all \( x \), the quadratic must equal zero, which leads us to:
\[ (\alpha + 1) = 0 \quad \text{and} \quad (1 - \alpha^2) = 0 \]

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

10. **Conclusion**:
The values of \( \alpha \) for which \( f(f(x)) = x \) are:
\[ \alpha = 1 \quad \text{or} \quad \alpha = -1 \]