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Question
Consider f : R → R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with inverse f−1 of f given by f−1 `(x)= sqrt (x-4)` where R+ is the set of all non-negative real numbers.
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Solution
Injectivity of f :
Let x and y be two elements of the domain (Q), such that
f(x) = f(y)
⇒ x2+4=y2+4
⇒ x2=y2
⇒ x = y (as co-domain as R+)
So, f is one-one
Surjectivity of f :
Let y be in the co-domain (Q), such that f(x) = y
⇒ x2 + 4 = y
⇒ x2 = y - 4
⇒ `x = sqrt (y-4) in R`
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Finding f -1:
Let f−1 (x) = y ...(1)
⇒ x = f (y)
⇒ x = y2 + 4
⇒ x − 4 = y2
⇒ ` y = sqrt(x-4)`
so, `f-1 (x) = sqrt(x-4)`
So , `f^-1 (x) = sqrt(x-4)` [from (1)]
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