Let `f: N->R` be a function defined as `f(x)=4x^2+12 x+15.` Show that `f: N->` Range `(f)` is invertible. Find the inverse of `fdot`

Explanation

`f : N rarr R`
`f(x) 4x^(2) + 12 x + 15`
`f(x_(1)) = f(x_(2))`
`rArr 4x_(1)^(2) + 12x_(1) + 15 = 4x_(2)^(2) + 12x_(2) + 15`
`rArr (4x_(1)^(2) - 4x_(2)^(2)) = 12(x_(2) - x_(1))`
`(x_(1)^(2) - x_(2)^(2)) = 12(x_(2) - x_(1))`
`(x_(1)^(2) - x_(2)^(2)) - 3(x_(1) - x_(2)) = 0`
`(x_(1) - x_(2))(x_(1) + x_(2) - 3) = 0`
`rArr x_(1) = x_(2)` Hence f is one-one
Now as `f : N rarr` range, so it has to be onto
Hence function is invertible i.e, `f^(-1)` exists
`f(x) = 4x^(2) + 12x + 15 = y`
`4x^(2) + 12x + 15 - y = 0`
`x = (-12 +- sqrt(144 - 16(15-y)))/(8)`
`= (-12 +- sqrt(16y - 96))/(8)`
`x = (-3 +- sqrt(y-6))/(2)`
`x = (-3 + sqrt(y-6))/(2)` is onto and `x = (-3-sqrt(y-6))/(2)` is not onto
`f^(-1) (y) = (-3+sqrt(y-6))/(2)`
`f^(-1) (x) = (-3+sqrt(x-6))/(2)`
`:. f^(-1)(x) = (sqrt(x-6)-3)/(2)`