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

Let `y` be an arbitrary element of range `f`.

Then `y = 4x^2 + 12x + 15`, for some `x in N`,

which implies that `y = (2x + 3)^2 + 6`.

This gives `x = (sqrt(y-6) - 3)/2`, as `y ge 6`.

Let us define `g : S → N` by `g (y) = (sqrt(y-6) - 3)/2`.

Now `gof (x) = g (f (x)) = g(4x^2 + 12x + 15) = g((2x + 3)^2 + 6)`

`=(sqrt((2x + 3)^2 + 6-6) - 3)/2`

`=(2x+3-3)/2`

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