If f(x) = frac{alpha x}{x + 1},x ne -1,for wh | vedclass

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(D) Given $f(x) = \frac{\alpha x}{x + 1}$.We need to find $f(f(x))$:$f(f(x)) = f\left( \frac{\alpha x}{x + 1} \right) = \frac{\alpha \left( \frac{\alp

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$-1$

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(D) Given $f(x) = \frac{\alpha x}{x + 1}$.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:$f(f(x)) = \frac{\frac{\alpha^2 x}{x + 1}}{\frac{\alpha x + x + 1}{x + 1}} = \frac{\alpha^2 x}{\alpha x + x + 1}$Given $f(f(x)) = x$,so:$\frac{\alpha^2 x}{(\alpha + 1)x + 1} = x$For this to hold for all $x$ (where $f$ is defined),we equate the coefficients or solve for $x 
eq 0$:$\alpha^2 x = x((\alpha + 1)x + 1)$$\alpha^2 x = (\alpha + 1)x^2 + x$For this to be an identity $f(f(x)) = x$,the coefficient of $x^2$ must be $0$ and the coefficient of $x$ must be $1$:$1) \alpha + 1 = 0 \implies \alpha = -1$$2) \alpha^2 = 1 \implies \alpha = \pm 1$Checking $\alpha = -1$:If $\alpha = -1$,then $f(f(x)) = \frac{(-1)^2 x}{(-1 + 1)x + 1} = \frac{x}{1} = x$.Thus,the correct value is $\alpha = -1$.

If $f(x) = \frac{\alpha x}{x + 1}$,$x \ne -1$,for what value of $\alpha$ is $f(f(x)) = x$?

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