Given as
f : R → R and f(x) = x2
(i) The domain of f = R (the set of real numbers)
As we know that the square of a real number is always positive or equal to zero.
∴ range of f = R+∪ {0}
(ii) Given as
f(x) = 4
As we know, x2 = 4
x2 – 4 = 0
(x – 2)(x + 2) = 0
∴ x = ± 2
∴ {x: f(x) = 4} = {–2, 2}
(iii) Given as
f(y) = –1
y2 = –1
However, the domain of f is R, and for every real number y, the value of y2 is non-negative.
Thus, there exists no real y for which y2 = –1.
∴ {y: f(y) = –1} = ∅