Let `f(x)=(alphax)/(x+1),x ne -1`. Then, for what values 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 values of \(\alpha\) such that \(f[f(x)] = x\) for the function \(f(x) = \frac{\alpha x}{x + 1}\), where \(x \neq -1\).

### Step-by-Step Solution:

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

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

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

5. **Set \(f[f(x)] = x\):**
We want:
\[ \frac{\alpha^2 x}{(\alpha + 1)x + 1} = x \]

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

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

8. **Factoring out \(x\):**
\[ x\left((\alpha + 1)x + (1 - \alpha^2)\right) = 0 \]
This gives us two cases: \(x = 0\) or \((\alpha + 1)x + (1 - \alpha^2) = 0\).

9. **Solving the quadratic equation:**
For the quadratic equation:
\[ (\alpha + 1)x + (1 - \alpha^2) = 0 \]
We can find \(x\):
\[ x = \frac{\alpha^2 - 1}{\alpha + 1} \]

10. **Setting the discriminant to zero:**
For \(f[f(x)] = x\) to hold for all \(x\), the discriminant must be zero:
\[ D = (1 - \alpha^2)^2 - 4(\alpha + 1)(0) = 0 \]
This leads to:
\[ 1 - \alpha^2 = 0 \implies \alpha^2 = 1 \implies \alpha = 1 \text{ or } \alpha = -1 \]

### Final Answer:
The values of \(\alpha\) for which \(f[f(x)] = x\) are:
\[ \alpha = 1 \quad \text{or} \quad \alpha = -1 \]