Your question is extremely unclear, but I'll have a go
f(N) = N + \frac{1}{N} \\ $ Let $y=f^{-1}(N)\\ N = y + \frac{1}{y} \\ yN = y^2 + 1 \\ y^2 - yN +1 = 0 \\ \\ $Using quadratic formula,$ \\ y = \frac{N \pm \sqrt{N^2 - 4}}{2} \\ \\ \therefore f^{-1}(N) = \frac{N \pm \sqrt{N^2 - 4}}{2}