For real x, let f(x)=x^3+5x+1, then:

Given $f(x)=x^3+5x+1$

Now $f'(x)=3x^2+5>0,∀\,x∈R$

∴ f (x) is strictly increasing function ∴ It is one-one

Clearly, f (x) is a continuous function also increasing on R,

$\underset{x→-∞}{\lim}f(x)=-∞$ and $\underset{x→∞}{\lim}f(x)=∞$

∴ f (x) takes every value between -∞ and ∞

Thus, f (x) is onto function.