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Let f(x)=(alphax)/(x+1),x!=-1. Then writ...

Let `f(x)=(alphax)/(x+1),x!=-1.` Then write the value of `alpha` satisfying `f(f(x))=x` for all `x!=-1.`

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We have,
`f{f(x)}=x ` for all `x !=0`
`=> f((alphax)/(x+1))=x` for all `x!=1`
`=> ((alpha((alphax))/(x+1))/(((alphax)/(x+1))+1)=x` for all `x!=0`
...
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