If a function f: R rightarrow R is defined by | vedclass

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(C) Given $f: R \rightarrow R$ such that $f(x)=x^3-x=x(x-1)(x+1)$.For one-one check: $f(1) = 1^3 - 1 = 0$ and $f(0) = 0^3 - 0 = 0$. Since $f(1) = f(0)

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onto but not one-one

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(C) Given $f: R ightarrow R$ such that $f(x)=x^3-x=x(x-1)(x+1)$.For one-one check: $f(1) = 1^3 - 1 = 0$ and $f(0) = 0^3 - 0 = 0$. Since $f(1) = f(0)$ but $1 
eq 0$,the function is not one-one.For onto check: $f(x)$ is a polynomial of odd degree. As $x ightarrow \infty$,$f(x) ightarrow \infty$ and as $x ightarrow -\infty$,$f(x) ightarrow -\infty$. Since the function is continuous,its range is $(-\infty, \infty)$,which is equal to the codomain $R$. Therefore,$f$ is onto.Thus,$f$ is onto but not one-one.

If a function $f: R \rightarrow R$ is defined by $f(x)=x^3-x$,then $f$ is

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