Assuming f(x) is invertible ∀x∈Rthen ∀k≠0,∃y=kx∋f−1(y)=kxi.e. f(kx) is invertible.Thus it must be the only value of k that makes the original statement is valid is k=0and the sum of this is just 0It's true that ∃f(0)∋f−1(f(0))=0 but f(0⋅x) is not invertible
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