Question

If f: R → R be the function defined by f(x) = 4x³ + 7, show that f is a bijection.

Answer 1

Given as f: R → R is a function defined by f(x) = 4x³ + 7

Injectivity:

Let x and y be any two elements in domain (R), such that f(x) = f(y)

⇒ 4x³ + 7 = 4y³ + 7

⇒ 4x³ = 4y³

⇒ x³ = y³

⇒ x = y

Therefore, f is one-one.

Surjectivity:

Let y be any element in co-domain (R), such that f(x) = y for some element x in R (domain)

f(x) = y

⇒ 4x³ + 7 = y

⇒ 4x³ = y − 7

⇒ x³ = (y – 7)/4

⇒ x = ³√(y - 7)/4 in R

Therefore, for every element in the co-domain, there exists some pre-image in the domain. f is onto.

Since, f is both one-to-one and onto, it is a bijection.