Consider `f: R^+->[-5,\ oo)` given by `f(x)=9x^2+6x+5` . Show that `f` is invertible with `f^(-1)(x)=(sqrt(x+6)-1)/3` .

Explanation

Given `f : R^(+) rarr [-5, oo)` given by `f(x) = 9x^(x^(2)) + 6x - 5` ...(i)
To test whether f is one-one : Let `x_(1), x_(2) in R^(+)`
Now `f(x_(1)) = f(x_(2)) rArr x_(1)^(2) + 6x_(1) - 5 = 9x_(2)^(2) + 6x_(2) - 5`
`rArr 9(x_(1)^(2)-x_(2)^(2)) + 6(x_(1) - x_(2)) = 0`
`rArr (x_(1)- x_(2)) [9(x_(1) + x_(2)) + 6] = 0 rArr x_(1) = x_(2)` So, f is one-one.
To test whether f is onto : Let `y in` co-domain `[-5, oo)`
Let `f(x) = y rArr 9x^(2)+6x - 5 = rArr 9x^(2) + 6x - (5+y) = 0`
`rArr x = (-6 +- sqrt(36 + 36(5+y)))/(18) = (-6 +- 6 sqrt(6+y))/(18) = (-1 +- sqrt(6+y))/(3)`
`x = (-1 + sqrt(6+y))/(3)` is onto
`x = (1- - sqrt(6+y))/(3)` is not onto `(because x in R^(+))`
Hence f is invertible i.e., `f^(-1)` exists.
To find `f^(-1) : y = f(x) rArr x = (sqrt(6+y)-1)/(3) rArr f^(-1)(y) = (sqrt(6+y)-1)/(3)`
`rArr f^(-1)(x) = (sqrt(6+x)-1)/(3), x in R^(+)`