Functions-20

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Discuss the surjectivity of the following functions:
(i) f : R → R given by f(x) = x3 + 2 for all x ∈ R
(ii) f : R → R given by f(x) = x2 + 2 for all x ∈ R.
(iii) f : Z → Z given by f(x) = 3x + 2 for all x ∈ Z.

Solution: (i) Let y be an arbitrary element of R. Then,

f(x) = y => x3 + 2 = y => x = (y - 2)1/3

Clearly, for all y ∈ R, (y - 2)1/3 is a real number. Thus, for all y ∈ R (co-domain) there exists

x = (y – 2)1/3 in R (domain) such that f(x) = x3 + 2 = y. Hence, f : R → R is an onto function.

(ii) Clearly, f(x) = x2 + 2 ≥ 2 for all x ∈ R. So, negative real numbers in R (domain) do not have their pre-

images in R (domain). Hence, f is not an onto function.

(iii) Let y be an arbitrary element of Z (co-domain). Then,

f(x) = y => 3x + 2 = y => x = (y - 2) / 3

Clearly, if y = 0, then x = -2/3 ∉ Z. Thus, y = 0 ∈ Z does not have its pre-image in Z(domain). Hence, f is not an onto

function.

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