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.

Answer 1

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)`