Question

If the function $f:[1,\infty )\to [1,\infty )$ is defined by $f\left(x\right)=2^{x(x-1)}$,then find $f^{-1}\left(4\right)$

Answer 1

$y=2^{x(x-1)}$
$lo{g}_2\left(y\right)=x^2-x$
$lo{g}_2\left(y\right)=(x-\frac{1}{2}{)}^2-\frac{1}{4}$
$\frac{4lo{g}_2\left(y\right)+1}{4}=(x-\frac{1}{2}{)}^2$
$x-\frac{1}{2}=\frac{\pm \sqrt{4lo{g}_2\left(y\right)+1}}{2}$
Since the range of f(x) is $R^{+}$
hence
$x-\frac{1}{2}=\frac{\sqrt{4lo{g}_2\left(y\right)+1}}{2}$
$x=\frac{1+\sqrt{4lo{g}_2\left(y\right)+1}}{2}$
Replacing x and y
$f^{-1}\left(x\right)=\frac{1+\sqrt{4lo{g}_2\left(x\right)+1}}{2}$
Therefore
$f^{-1}\left(4\right)=\frac{1+\sqrt{8+1}}{2}$
$=\frac{4}{2}$
$=2$