Check the injectivity and surjectivity of the | vedclass

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(A) The function is $f: Z \rightarrow Z$ defined by $f(x) = x^{3}$.For injectivity: Let $x, y \in Z$ such that $f(x) = f(y)$.Then $x^{3} = y^{3}$. Sin

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Injective but not surjective

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(A) The function is $f: Z ightarrow Z$ defined by $f(x) = x^{3}$.For injectivity: Let $x, y \in Z$ such that $f(x) = f(y)$.Then $x^{3} = y^{3}$. Since the cube root function is well-defined for all real numbers,$x^{3} = y^{3} \implies x = y$.Therefore,$f$ is injective.For surjectivity: $A$ function is surjective if for every $y \in Z$,there exists an $x \in Z$ such that $f(x) = y$.Consider $y = 2 \in Z$. We need $x^{3} = 2$,which implies $x = \sqrt[3]{2}$.Since $\sqrt[3]{2} 
otin Z$,there is no element $x \in Z$ such that $f(x) = 2$.Therefore,$f$ is not surjective.Conclusion: The function $f$ is injective but not surjective.

Check the injectivity and surjectivity of the function $f: Z \rightarrow Z$ defined by $f(x) = x^{3}$.

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