Let $f(x)=x^2+\frac{1}{x^2}$ and $g(x)=x-\frac{1}{x}$ for $x \in R-\{-1,0,+1\}$, then the local minimum of…
Let $f(x)=x^2+\frac{1}{x^2}$ and $g(x)=x-\frac{1}{x}$ for $x \in R-\{-1,0,+1\}$, then the local minimum of $\frac{f(x)}{g(x)}$ is
$-3$
$2 \sqrt{2}$
$-2 \sqrt{2}$
$3$
Solution
$f(x)=x^2+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)+2, g(x)=x-\frac{1}{x}$
Let $x-\frac{1}{x}=t$
$f(x)=t^2+2, g(x)=t$
$\begin{aligned} & \frac{f(x)}{g(x)}=t+\frac{2}{t}=h(t) \\ & h^{\prime}(t)=1-\frac{2}{t^2}\end{aligned}$
On putting $h^{\prime}(t)=0$
$\begin{aligned} \Rightarrow & & 1-\frac{2}{t^2} & =0 \\ \Rightarrow & & t & = \pm \sqrt{2} \\ \Rightarrow & & h^{\prime}(t) & =\frac{4}{t^3}\end{aligned}$
$h^{\prime \prime}(-\sqrt{2}) < 0$ and $h^{\prime \prime}(\sqrt{2})>0$
$\therefore t=\sqrt{2}$ is a point of minima. Local minimum value of $h(t)=\sqrt{2}+\frac{2}{\sqrt{2}}=2 \sqrt{2}$
$\therefore$ Required local minimum value
$=2 \sqrt{2}$