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

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Question

(d) $f(x) = \frac{{\alpha x}}{{x + 1}}$;

$f(f(x)) = f\left( {\frac{{\alpha x}}{{x + 1}}} \right) = \frac{{\alpha \left( {\frac{{\alpha x}}{{x + 1}}

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(d) $f(x) = \frac{{\alpha x}}{{x + 1}}$;

$f(f(x)) = f\left( {\frac{{\alpha x}}{{x + 1}}} ight) = \frac{{\alpha \left( {\frac{{\alpha x}}{{x + 1}}} ight)}}{{\frac{{\alpha x}}{{x + 1}} + 1}}$

But $f(f(x)) = x$, 

$	herefore$ $\frac{{{\alpha ^2}x}}{{\alpha x + x + 1}} = x$

Put $\alpha = - 1$, $\frac{{{{( - 1)}^2}x}}{{( - 1)x + x + 1}} = \frac{x}{{ - x + x + 1}} = x$; 

$	herefore$ $\alpha = - 1$.

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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