Q. 11 D5.0( 1 Vote )

Let f be a real function given by . Find each of the following:

f2

Also, show that fof ≠ f2.

Answer :

We have, f(x) =


Clearly, domain of f = [2, ∞] and range of f = [0, ∞)


We observe that range of f is not a subset of domain of f


Domain of (fof) = {x: x ϵ Domain of f and f(x) ϵ Domain of f}


= {x: x ϵ [2, ∞) and ϵ [2, ∞)}


= {x: x ϵ [2, ∞) and ≥ 2}


= {x: x ϵ [2, ∞) and x – 2 ≥ 4}


= {x: x ϵ [2, ∞) and x ≥ 6}


= [6, ∞)


Now,


(fof)(x) = f(f(x)) = f =


fof: [6, ∞) R defined as


(fof)(x) =


f2(x) = [f(x)]2 = = x – 2


f2: [2, ∞) R defined as


f2(x) = x – 2


fof ≠ f2


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