We have, $f\left(x\right)=x+\frac{1}{x},x\ne 0$
On differentiating w.r.t.x, we get
$f\text{'}\left(x\right)=1-\frac{1}{x^2}f"\left(x\right)=$
For maxima or minima, we put $f^{\text{'}}\left(x\right)=0$
$\Rightarrow 1-\frac{1}{x^2}=0$
$\Rightarrow x^2-1=0\Rightarrow x=-1,1$
Now, since $f^{\text{'}}\left(x\right)>0,x\in \left(-1-h,-1\right)$ and $f^{\text{'}}\left(x\right)<0,x\in \left(-1,-1+h\right)$

$\Rightarrow x=-1$ is point of local maxima.
and local maxima value is $f\left(-1\right)=-1+\frac{1}{\left(-1\right)}=-2$
and $f^{\text{'}}\left(x\right)<0$ when $x\in \left(1-h,1\right)$
and $f”\left(x\right)>0$ when $x\in \left(\mathrm{1,1}+h\right)$
$\Rightarrow x=1$ Is point of minima.
and local minima value, $f\left(1\right)=1+\frac{1}{1}=2$