Q. 115.0( 3 Votes )

A function f : R

Answer :

We have f : R R and f(x) = x3 + 4.


Recall that a function is a bijection only if it is both one-one and onto.


First, we will check if f is one-one.


Let x1, x2ϵ R (domain) such that f(x1) = f(x2)


x13 + 4 = x23 + 4


x13 = x23


(x1 – x2)(x12 + x1x2 + x22) = 0


As x1, x2ϵ R and the second factor has no real roots,


x1 – x2 = 0


x1 = x2


So, we have f(x1) = f(x2) x1 = x2.


Thus, function f is one-one.


Now, we will check if f is onto.


Let y ϵ R (co-domain) such that f(x) = y


x3 + 4 = y


x3 = y – 4



Clearly, for every y ϵ R, there exists x ϵ R (domain) such that f(x) = y and hence, function f is onto.


Thus, the function f is a bijection and has an inverse.


We have f(x) = y x = f-1(y)


But, we found f(x) = y


Hence,


Thus, f(x) is invertible and


Hence, we have



Thus, f-1(3) = –1.


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