First slide

Functions (XII)

Question

If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ then $f^{- 1} (x)$ is

Moderate

A

$(1 / 2)^{x (x - 1)}$
B

$\frac{1}{2} (1 + \sqrt{1 + 4 \log_{2} x})$
C

$\frac{1}{2} (1 - \sqrt{1 + 4 \log_{2} x})$
D

not defined

Solution

Let $f (x) = y = 2^{x (x - 1)}$
$\begin{array}{l} \Rightarrow \log y = x (x - 1) \log 2 \\ \Rightarrow x^{2} - x - \log_{2} y = 0 \\ \Rightarrow x = \frac{1 \pm \sqrt{1 + 4 \log_{2} y}}{2} \\ \Rightarrow x = f^{- 1} (y) = \frac{1 \pm \sqrt{1 + 4 \log_{2} y}}{2} \end{array}$
$\Rightarrow f^{- 1} (x) = \frac{1 + \sqrt{1 + 4 \log_{2} y}}{2}$ as $f^{- 1} (x) \geq 1$ .

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