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, f : R → R and f(x) = x². i. range of f Domain of f = R (set of real numbers) We know that the square of a real number is always positive or equal to zero. Hence, The range of f is the set of all non-negative real numbers. Thus, Range of f = R +∪ {0} ii. {x: f(x) = 4} Given, f(x) = 4 ⇒ x²= 4 ⇒ x² – 4 = 0 ⇒ (x – 2)(x + 2) = 0 ∴ x = ±2 Thus, {x: f(x) = 4} = {–2, 2} iii. {y: f(y) = –1} Given, f(y) = –1 ⇒ y² = –1 However, The domain of f is R, and for every real number y, the value of y²is non-negative. Hence, There exists no real y for which y² = –1. Thus, {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,

f : R → R and f(x) = x².

i. range of f

Domain of f = R

(set of real numbers)

We know that the square of a real number is always positive or equal to zero.

Hence,

The range of f is the set of all non-negative real numbers.

Thus,

Range of f = R +∪ {0}

ii. {x: f(x) = 4}

Given,

f(x) = 4

⇒ x²= 4

⇒ x² – 4 = 0

⇒ (x – 2)(x + 2) = 0

∴ x = ±2

Thus,

{x: f(x) = 4} = {–2, 2}

iii. {y: f(y) = –1}

Given,

f(y) = –1

⇒ y² = –1

However,

The domain of f is R, and for every real number y, the value of y²is non-negative.

Hence,

There exists no real y for which y² = –1.

Thus,

{y: f(y) = –1} = ∅