Find the Value of \( a \)
Let
\[ f(x)=\frac{ax}{x+1}, \qquad x\ne-1 \]
Then write the value of \(a\) satisfying
\[ f(f(x))=x \]
for all \(x\ne-1\).
Given,
\[ f(x)=\frac{ax}{x+1} \]
Therefore,
\[ f(f(x)) = \frac{a\left(\frac{ax}{x+1}\right)} {\frac{ax}{x+1}+1} \]
\[ = \frac{a^2x}{(a+1)x+1} \]
Since
\[ f(f(x))=x \]
we get
\[ \frac{a^2x}{(a+1)x+1}=x \]
Comparing coefficients,
\[ a+1=0 \]
\[ a=-1 \]
Therefore,
\[ \boxed{a=-1} \]