Question

Let $f(x)=\left|\begin{array}{ccc}x& 1& \frac{-3}{2}\\ 2& 2& 1\\ \frac{1}{x-1}& 0& \frac{1}{2}\end{array}\right|$ The minimum value of $f(x)$ (given $x>1$ ), is

Answer 1

Answer: 4

$y=\left|\begin{array}{ccc}x& 1& \frac{-3}{2}\\ 2& 2& 1\\ \frac{1}{x-1}& 0& \frac{1}{2}\end{array}\right|$
$=x(1-0)-\left(1-\frac{1}{x-1}\right)-\frac{3}{2}\left(0-\frac{2}{x-1}\right)$
$=x-1+\frac{1}{x-1}+\frac{3}{x-1}$
$=(x-1)+\frac{4}{(x-1)}$
By applying A.M.-G.M. inequality, we have
$y=(x-1)+\frac{4}{x-1}\ge 2\sqrt{(x-1)\frac{4}{x-1}}=4$ (As $x>1$)