Question

If f:[1,∞) → [2,∞) is given by f(x) = x+1/x find f^-1(x) , {assume bijection)

Answer 1

Let y = f(x)

$therefore \: y = \frac{ {x}^{2} + 1 }{x} = > {x}^{2} - xy + 1 = 0 \\ = > x = \frac {y + _{ - } \sqrt{ {y}^{2} - 4} }{2 } \\ = > f ^{ - 1} (y) = \frac{y + _{ - } \sqrt{ {y}^{2} - 4 } }{2} \: \: \: ( as \: f(x) = y = > x = f ^{ - 1} (y) \\ \\ = > f ^{ - 1}(x) = \frac{x + _{ - } \sqrt{ {x}^{2} - 4 } }{2}$

Since the range of inverse function is [1,∞), therefore neglecting the negative sign, we have

$f ^{ - 1} (x) = \frac{x + \sqrt{ {x}^{2} - 4 } }{2}$