Given that \[f\left( x \right) = {x^3} + 5x + 1\]
\[\therefore\] \[\,\,\,\,\,\] \[f'\left( x \right) = 3{x^2} + 5 > 0,\,\,\,\forall x \in R\]
\[ \Rightarrow f\left( x \right)\,\,\] is strictly increasing on R
\[ \Rightarrow f\left( x \right)\] is one one
\[\therefore\] \[\,\,\,\,\,\,\,\] Being a polynomial f(x) is cont. and inc.
on R with \[\mathop {\lim }\limits_{x \to \infty } \,f\left( x \right) = - \infty \]
and \[\mathop {\lim }\limits_{x \to \infty } \,f\left( x \right) = \infty \]
\[\therefore\] \[\,\,\,\,\,\,\,\] Range of \[f = \left( { - \infty ,\infty } \right) = R\]
Hence f is onto also, So, f is one and onto R.