Write the minimum value

Given : $f(x)=x+\frac{1}{x}$

$\Rightarrow f^{\prime}(x)=1-\frac{1}{x^{2}}$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

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

$\Rightarrow x^{2}=1$

$\Rightarrow x=1,-1$

But $x>0$

$\Rightarrow x=1$

Now,

$f^{\prime \prime}(x)=\frac{1}{x^{3}}$

At $x=1:$

$f^{\prime \prime}(1)=\frac{2}{(1)^{3}}=2>0$

So, $x=1$ is a point of local minimum.

Thus, the local minimum value is given by

$f(1)=1+\frac{1}{1}=1+1=2$