If f: R → R be the function defined by f(x) = 4x3 + 7, show t

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

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

Answer»

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.

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