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.

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 Rate this question :

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