Question

Consider f: R+$\to$[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+ $\to$ [4, ∞) is given as f(x) = x2 + 4.

**One-one: **
Let f(x) = f(y).
$\implies$x2+4 = y2+4
$\implies$ x2 = y2
$\implies$x = y [as x = y ∈ R+]
∴ f is a one-one function.

Onto:
For y ∈ [4, ∞), let y = x2+ 4.
$\implies$x2 = y-4 ≥ 0 [as y ≥ 4]
$\implies$ 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) = $\sqrt {(x2+4)-4}$ = $\sqrt {x^2}$ = x
And fog(y) = f(g(y)) = f$(\sqrt {y-4})$= $\sqrt {(y-4)^2-4}$ = (y - 4) + 4 = y.
therefore gof = fog = IR+

Hence, f is invertible and the inverse of f is given by
f-1 = g(y) = $\sqrt {y-4}$