Question

Consider f: R+$\to$ [−5,∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with
$f^{-1}(y) = \frac {(\sqrt {y+6})-1}{3}$

Answer 1

f: R+$\to$ [−5, ∞) is given as f(x) = 9x2+6x−5.
Let y be an arbitrary element of [−5, ∞).
Let y = 9x2+6x−5.
y=(3x+1)2-1-5 = (3x+1)2-6
$\implies$(3x+1)2 = y+6
$\implies$3x+1 = $\sqrt {y+6}$
$\implies$x = $\frac {\sqrt {y+6}-1}{3}$
∴f is onto, thereby range f = [−5, ∞).
Let us define g: [−5, ∞) $\to$ R+ as g(y) =$\frac {\sqrt {y+6}-1}{3}$
We now have:
(gof)(x) = g(f(x)) = g(9x2+6x-5)
=g((3x+1)2-6)
=$\frac {\sqrt {(3x+1)2-6+6}-1}{3}$
=$\frac {3x+1-1}{3}$
=x
And (fog)(y) = f(g(y) = f$(\frac {\sqrt {y+6}-1}{3})$
=$[3(\frac {\sqrt {y+6}-1}{3})+1]^2-6$
=$(\sqrt {y+6})^2 -6$ = y+6-6 = y.
therefore gof=IR and fog =I[-5,∞].
Hence, f is invertible and the inverse of f is given by
f-1(y) = g(y) =$\frac {\sqrt {y+6}-1}{3}$