First slide

Functions (XII)

Question

If $f : [1, \infty) \to [2, \infty)$ is given by $f (x) = x + \frac{1}{x}$ then $f^{- 1} (x)$ equals

Moderate

A

$\frac{(x + \sqrt{x^{2} - 4})}{2}$
B

$\frac{x}{1 + x^{2}}$
C

$\frac{(x - \sqrt{x^{2} - 4})}{2}$
D

$1 + \sqrt{x^{2} - 4}$

Solution

$\begin{array}{l} f : [1, \infty) \to [2, \infty) \\ f (x) = x + \frac{1}{x} = y \\ or x^{2} - y x + 1 = 0 \\ or x = \frac{y \pm \sqrt{y^{2} - 4}}{2} \\ But given f : [1, \infty) \to [2, \infty) \\ ∴ x = \frac{y + \sqrt{y^{2} - 4}}{2} \\ or f^{- 1} (x) = \frac{x + \sqrt{x^{2} - 4}}{2} \end{array}$

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