The value of parameter `alpha`, for which the function `f(x) = 1+alpha x, alpha!=0` is the inverse of itself

Explanation

Let `f(x)=y`, then `1+alpha x=y`



`Rightarrow {x}=frac{{y}-1}{alpha} `


`therefore {f}^{-1}({y})=frac{{y}-1}{alpha}`


`therefore {f}^{-1}({x})=frac{{x}-1}{alpha}`



`{Now} f({x})={f}^{-1}({x}) Rightarrow 1+alpha {x}=frac{{x}-1}{alpha} Rightarrow 1+alpha {x}=frac{1}{alpha} {x}-frac{1}{alpha} ldots ldots . {As} {f}({x})` is invertible function itself



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