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Question
Consider f: R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by `f^(-1)(y) = sqrt(y - 4)`, where R+ is the set of all non-negative real numbers.
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Solution
f: R+ → [4, ∞) is given as f(x) = x2 + 4.
One-one:
Let f(x) = f(y).
⇒ x2 + 4 = y2 + 4
⇒ x2 = y2
⇒ x = y ...[As x = y ∈ R+]
∴ f is a one-one function.
Onto:
For y ∈ [4, ∞), let y = x2 + 4.
⇒ x2 = y – 4 > = 0 ...[As y ≥ 4]
⇒ `x = sqrt(y - 4) ≥ 0`
Therefore, for any y ∈ R, there exists `x = sqrt(y - 4) ∈ R` such that
`f(x) = f(sqrt(y - 4))`
= `(sqrt(y - 4))^2 + 4`
= y – 4 + 4
= y
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define g: [4, ∞) → R+ by,
`g(y) = sqrt(y - 4)`
Now, gof(x) = g(f(x))
= g(x2 + 4)
= `sqrt((x^2 + 4) - 4)`
= `sqrt(x^2)`
= x
And fog(y) = f(g(y))
= `f(sqrt(y - 4))`
= `(sqrt(y - 4))^2 + 4`
= (y – 4) + 4
= y
∴ gof = fog = IR+
Hence, f is invertible and the inverse of f is given by`f^(-1)(y) = g(y) = sqrt(y - 4)`.
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