Question

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

• A. $3$
• B. $-3$
• C. $-2\sqrt{2}$
• D. $\sqrt[2]{22}$

Answer 1

The correct option is D $\sqrt[2]{22}$

$f\left(x\right)=x^2+\frac{1}{x^2}$

$g\left(x\right)=x-\frac{1}{x}$

$f\left(x\right)=x^2+\frac{1}{x^2}$

$f\left(x\right)={(x-\frac{1}{x})}^2+2$

Let $x-\frac{1}{x}=t=g\left(x\right)$

$f\left(x\right)=t^2+2$

$\begin{array}{cc}h\left(x\right)=K\left(t\right)& =\frac{t^2+2}{t}\\ & =t+\frac{2}{t}\end{array}$

$k^{\prime }\left(t\right)=1-\frac{2}{t^2}$

$k^{\prime }\left(t\right)=0$

$\Rightarrow 1-\frac{2}{t^2}=0$

$\Rightarrow t^2=2$

$\therefore t=\pm \sqrt{2}$

$K^{\prime \prime }\left(t\right)=\frac{4}{t^3}$

At $t=\sqrt{2},K^{\prime \prime }\left(t\right)=\frac{4}{2\sqrt{2}}>0$

Which shows minimum value of $h\left(x\right)$

$K^{\prime \prime }\left(t\right)=\frac{4}{2\sqrt{2}}=\sqrt{2}$