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

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : R → R, defined by f(x) = 4x³ + 7

To Prove : – f : R → R is bijective defined by f(x) = 4x³ + 7

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

⇒ f(x) = f(y)

⇒ 4x³ + 7 = 4y³ + 7

⇒ x³ = y³

⇒ x = y

Hence, f is One – One function

Check for Surjectivity:

Let y be element belongs to R i.e be arbitrary, then

⇒ f(x) = y

⇒ 4x³ + 7 = y

⇒ 4x³ + 7 – y = 0

Now, we know that for 3 degree equation has a real root

So, let be that root

⇒

⇒

Thus for clearly , there exist such that f(x) = y

Therefore f is onto

Thus, It is Bijective function

Hence Proved