Question: Given a one-to-one function and a not one-to-one function, is it possible that the not one-to-one function composed with the one-to-one function is one-to-one. Show

Given a one-to-one function and a not one-to-one function, is it possible that the not one-to-one function composed with the one-to-one function is one-to-one. Show this using f :R ->R, f(x) = e^x and g : R -> R, g(x) = x^2.

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