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Archangel Macsika

whether each of these functions is a bijection from R to R. f (x) = x 3 -1

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Consider the function f:\R\to\R,\ f(x)=x^3-1. Let us show that this function is an injection. Let f(x)=f(y). Then x^3-1=y^3-1, and hence x^3=y^3. It follows that x=y, and f is an injective function. Further, let us show that this function is a surjection. Since for any y\in\R the equation f(x)=y is equivalent to x^3-1=y, and hence has the solution x=\sqrt[3]{y+1}\in\R, we conclude that f is a surjection. Consequently, the function f is a bijection.


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Question ID: mtid-5-stid-8-sqid-2649-qpid-1119