Let f(x) = frac{alpha x}{x+1},x neq -1. If f( | vedclass

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(B) Given the function $f(x) = \frac{\alpha x}{x+1}$.We need to find $f(f(x))$:$f(f(x)) = f\left(\frac{\alpha x}{x+1}\right) = \frac{\alpha \left(\fra

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$-1$

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(B) Given the function $f(x) = \frac{\alpha x}{x+1}$.We need to find $f(f(x))$:$f(f(x)) = f\left(\frac{\alpha x}{x+1}ight) = \frac{\alpha \left(\frac{\alpha x}{x+1}ight)}{\frac{\alpha x}{x+1} + 1}$Simplify the expression:$f(f(x)) = \frac{\frac{\alpha^2 x}{x+1}}{\frac{\alpha x + x + 1}{x+1}} = \frac{\alpha^2 x}{(\alpha + 1)x + 1}$Given $f(f(x)) = x$,so:$\frac{\alpha^2 x}{(\alpha + 1)x + 1} = x$Assuming $x 
eq 0$,we divide by $x$:$\frac{\alpha^2}{(\alpha + 1)x + 1} = 1$$\alpha^2 = (\alpha + 1)x + 1$For this to hold for all $x$ in the domain,the coefficient of $x$ must be $0$ and the constant term must be $1$:$\alpha + 1 = 0 \implies \alpha = -1$Checking the constant term: $(-1)^2 = 1$,which is true.Thus,$\alpha = -1$.

Let $f(x) = \frac{\alpha x}{x+1}$,$x \neq -1$. If $f(f(x)) = x$,then the value of $\alpha$ is . . . . . . .

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