For real x, let $f\left(x\right)=x^3+5x+1$, then

The function $f:R\to R$ is defined as $f\left(x\right)=x^3+5x+1$

Let $y\in R$ then $y=x^3+5x+1\Rightarrow x^3+5x+1-y=0$

As a polynomial of odd degree has always at least one real root, corresponding to any $y\in$co-domain.

So $\exists$ some $x\in$ domain such that $f\left(x\right)=y$. Hence f is onto

Also f is continuous on R, because its a polynomial function. Now $f\text{'}\left(x\right)=3x^2+5>0$

$\therefore f$ is strictly increasing. Hence f is one-one also