Check the injectivity and surjectivity of the function $f: R \rightarrow R$ defined by $f(x) = x^2$.
(NONE) The function $f: R \rightarrow R$ is defined by $f(x) = x^2$.
For injectivity:
Consider $f(-1) = (-1)^2 = 1$ and $f(1) = (1)^2 = 1$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function is not injective (one-to-one).
For surjectivity:
The codomain of $f$ is $R$ (the set of all real numbers).
For any $x \in R$,$x^2 \geq 0$. Thus,negative real numbers in the codomain have no pre-image in the domain.
For example,$-2 \in R$,but there exists no $x \in R$ such that $f(x) = -2$ (as $x^2 = -2$ has no real solution).
Therefore,the function is not surjective (onto).
Conclusion: The function $f$ is neither injective nor surjective.
For injectivity:
Consider $f(-1) = (-1)^2 = 1$ and $f(1) = (1)^2 = 1$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function is not injective (one-to-one).
For surjectivity:
The codomain of $f$ is $R$ (the set of all real numbers).
For any $x \in R$,$x^2 \geq 0$. Thus,negative real numbers in the codomain have no pre-image in the domain.
For example,$-2 \in R$,but there exists no $x \in R$ such that $f(x) = -2$ (as $x^2 = -2$ has no real solution).
Therefore,the function is not surjective (onto).
Conclusion: The function $f$ is neither injective nor surjective.
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