Show that the function f: R rightarrow R defi | vedclass

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(N/A) To check for one-one: $A$ function $f$ is one-one if $f(x_{1}) = f(x_{2})$ implies $x_{1} = x_{2}$.Here,$f(1) = (1)^{2} = 1$ and $f(-1) = (-1)^{

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(N/A) To check for one-one: $A$ function $f$ is one-one if $f(x_{1}) = f(x_{2})$ implies $x_{1} = x_{2}$.Here,$f(1) = (1)^{2} = 1$ and $f(-1) = (-1)^{2} = 1$.Since $f(1) = f(-1)$ but $1 
eq -1$,the function $f$ is not one-one.To check for onto: $A$ function $f: R ightarrow R$ is onto if for every $y \in R$ (co-domain),there exists an $x \in R$ (domain) such that $f(x) = y$.Here,the range of $f(x) = x^{2}$ is $[0, \infty)$,which is a subset of the co-domain $R$.For example,consider $y = -2$ in the co-domain $R$. There is no real number $x$ such that $x^{2} = -2$,because the square of any real number is non-negative.Therefore,$f$ is not onto.

Show that the function $f: R \rightarrow R$ defined as $f(x) = x^{2}$ is neither one-one nor onto.

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