Given that f(x)=x3+5x+1 ∴f′(x)=3x2+5>0,∀x∈R ⇒f(x) is strictly increasing on R ⇒f(x) is one one ∴ Being a polynomial f(x) is cont. and inc. on R with
lim
x→∞
f(x)=−∞ and
lim
x→∞
f(x)=∞ ∴ Range of f=(−∞,∞)=R Hence f is onto also, So, f is one and onto R.