Answer :

It is given that f : R+ [4, ) given by f (x) = x2 + 4.

Now, Let f(x) = f(y)


x2 + 4 = y2 + 4


x2 = y2


x = y


f is one-one function.


Now, for y ϵ [4, ∞), let y = x2 + 4.


x2 = y -4 ≥ 0



for any y ϵ R, there exists x = ϵ R such that


= y -4 + 4 = y.


f is onto function.


Therefore, f is one–one and onto function, so f-1 exists.


Now, let us define g: [4, ∞) R+ by,


g(y) =


Now, gof(x) = g(f(x)) = g(x2 + 4) =


And, fog(y) = f(g(y)) = =


Therefore, gof = gof = IR.


Therefore, f is invertible and the inverse of f is given by


f-1(y) = g(y) =


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