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

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) = Rate this question :

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