Q. 64.1( 8 Votes )

# A function f : R → R is defined by f(x) = x^{2}. Determine

i. range of f

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

iii. {y: f(y) = –1}

Answer :

Given f : R → R and f(x) = x^{2}.

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

⇒ x^{2} = 4

⇒ x^{2} – 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

⇒ y^{2} = –1

However, the domain of f is R, and for every real number y, the value of y^{2} is non-negative.

Hence, there exists no real y for which y^{2} = –1.

Thus, {y: f(y) = –1} = ∅

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