Q. 94.0( 4 Votes )

Consider f : R+ [–5, ∞) given by f(x) = 9x2 + 6x – 5. Show that f is invertible with.

Answer :

We have f : R+ [–5, ∞) and f(x) = 9x2 + 6x – 5.


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)


9x12 + 6x1 – 5 = 9x22 + 6x2 – 5


9x12 + 6x1 = 9x22 + 6x2


9x12 – 9x22 + 6x1 – 6x2 = 0


9(x12 – x22) + 6(x1 – x2) = 0


9(x1 – x2)(x1 + x2) + 6(x1 – x2) = 0


(x1 – x2)[9(x1 + x2) + 6] = 0


x1 – x2 = 0 (as x1, x2ϵ R+)


x1 = x2


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 [–5, ∞) (co-domain) such that f(x) = y


9x2 + 6x – 5 = y


Adding 6 to both sides, we get


9x2 + 6x – 5 + 6 = y + 6


9x2 + 6x + 1 = y + 6


(3x + 1)2 = y + 6





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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