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=I_R and fog =I_[-5,∞].
Hence, f is invertible and the inverse of f is given by 
f^-1(y) = g(y) =\(\frac {\sqrt {y+6}-1}{3}\)