A function f: R → R is defined by f(x) = x2. Determine(i) range of

1.	A function f: R → R is defined by f(x) = x2. Determine(i) range of f(ii) {x: f(x) = 4}(iii) {y: f(y) = –1}
Answer» 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} = ∅

A function f: R → R is defined by f(x) = x2. Determine(i) range of f(ii) {x: f(x) = 4}(iii) {y: f(y) = –1}

Answer»

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} = ∅