Question

If $f\left(x\right)=x^2+\frac{1}{x^2}\text{and}g\left(x\right)=x-\frac{1}{x},xϵR-\{-1,0,1\},\text{and}h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$ then the local minimum value of $h\left(x\right)$ is

• A. $-2\sqrt{2}$
• B. -3
• C. $2\sqrt{2}$
• D. 3

Answer 1

The correct option is C $2\sqrt{2}$

$h\left(x\right)=\frac{t^2+2}{t},\text{where}t=x-\frac{1}{x}=t+\frac{2}{t}$
$h^{\prime }\left(x\right)=1-\frac{2}{t^2}=0\Rightarrow t=\pm \sqrt{2}$
$h^{\prime \prime }\left(x\right)=\frac{4}{t^3}$
Local minimum at $t=\sqrt{2}$
$h\left(\sqrt{2}\right)=\sqrt{2}+\frac{2}{\sqrt{2}}=2\sqrt{2}$