The function f: R [0, ) defined by f(x) = x^2+1 - x is :

Step-by-step solution

f(x) = x^2+1 - x f'(x) = x x^2+1 - 1 = x - x^2+1 x^2+1 Since x^2+1 > x^2 = |x| x for all x R , we have x - x^2+1 Thus, f'(x) To find the range, we evaluate the limits at infinity: _ x - f(x) = _ x - ( x^2+1 - x) = _ x f(x) = _ x ( x^2+1 - x) = _ x 1 x^2+1 + x = 0 Since f(x) is continuous and strictly decreasing, its range is (0, ) . The given codomain is [0, ) . Since 0 is not in the range, f(x) is not onto. Hence, the function is one-one but not onto. Answer: One-one but not onto