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


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