If a function $f: \mathrm{R} \rightarrow \mathrm{R}$ is defined by $f(x)=x^3-x$, then $f$ is
If a function $f: \mathrm{R} \rightarrow \mathrm{R}$ is defined by $f(x)=x^3-x$, then $f$ is
one-one and onto
one-one but not onto
onto but not one-one
neither one-one nor onto
Solution
Given $f: \mathrm{R} \rightarrow \mathrm{R}$ such that $f(x)=x^3-x=x(x-1)(x+1)$
$\because f(1)=0=f(0)$. So, $f(x)$ is not one-one Since, $f(x)=x^3-x$ is a polynomial function
So it is continuous on R and, If $x \rightarrow \infty \Rightarrow f(x) \rightarrow \infty$
and $x \rightarrow-\infty \Rightarrow f(x) \rightarrow-\infty$. So, range of $f(x)$ is $(-\infty, \infty)$ $\Rightarrow f(x)$ is an onto function