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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.
Solution
f: R+ → [4, ∞) is given as f(x) = x2 + 4.
One-one:
Let f(x) = f(y).
`=> x^2 + 4 = y^2 + 4`
`=> x^2 = y^2`
=> x = y [as `x = y in R`]
∴ f is a one-one function.
Onto:
For y ∈ [4, ∞), let y = x2 + 4.
`=> x^2 = y - 4 >= 0 ` [as `y >= 4`]
`=> x = sqrt(y-4) >= 0`
Therefore, for any y ∈ R, there exists `x = sqrt(y - 4) in 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(x^2 + 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= I_(R+)`
Hence, f is invertible and the inverse of f is given by
`f^(-1) = g(y) = sqrt(y - 4)`
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