Question

For real x, let f (x) = x³ + 5x + 1, then
(a) f is onto R but not one-one
(b) f is one-one and onto R
(c) f is neither one-one nor onto R
(d) f is one-one but not onto R

Answer 1

(b) : The function is f : R → R
f(x) = x³ + 5x + 1
Let y ∈ R then y = x³ + 5x + 1
⇒ x³ + 5x + 1 – y = 0
As a polynomial of odd degree has always at least one real root, corresponding to any y ∈ co-domain there ∃ some x ∈ domain such that f(x) = y. Hence f is onto .
Also f is continuous on R, because it’s a polynomial function f ′(x) = 3x² + 5 > 0
∴ f is strictly increasing
Hence f is one-one also.