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 ⇒
Thus, f(x) is invertible and
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