The function f : R → R defined as f x = x 2 - x + 1 x 2 + x + 1 is

Step-by-step solution

f x = x 2 − x + 1 x 2 + x + 1 ⇒ f ' x = 2 x 2 − 1 x 2 + x + 1 2 f ' x is positive in - ∞ , - 1 ∪ 1 , ∞ and negative in - 1 , 1 Hence, f x is many-one. For Range of f x , let y = x 2 - x + 1 x 2 + x + 1 ⇒ y - 1 x 2 + y + 1 x + y - 1 = 0 ⇒ since x ∈ R , D ≥ 0 ⇒ y + 1 2 - 4 y - 1 2 ≥ 0 ⇒ - 3 y 2 + 10 y - 3 ≥ 0 ⇒ y ∈ 1 3 , 3 Hence f x is into