Q. 75.0( 1 Vote )

Consider f : R+ [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with f-1 of f given by, where R+ is the set of all non-negative real numbers.

Answer :

We have f : R+ [4, ∞) and f(x) = x2 + 4.


Recall that a function is invertible only when it is both one-one and onto.


First, we will prove that f is one-one.


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


x12 + 4 = x22 + 4


x12 = x22


x1 = x2 (x1≠–x2as x1, x2ϵ R+)


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


Thus, function f is one-one.


Now, we will prove that f is onto.


Let y ϵ [4, ∞) (co-domain) such that f(x) = y


x2 + 4 = y


x2 = y – 4



Clearly, for every y ϵ [4, ∞), there exists x ϵ R+ (domain) such that f(x) = y and hence, function f is onto.


Thus, the function f 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


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