Question
Write the minimum value of $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}},\text{x}>0.$

Answer

Given, $\text{f}\text{(x)}=\text{x}+\frac{1}{\text{x}}$
Implies that $\text{f}'\text{(x)}=1-\frac{1}{\text{x}^2}$
For a local maxima or a local minima, We must have 
$\text{f}'\text{(x)}=0$
Implies that $1-\frac{1}{\text{x}^2}=0$
Implies that $\text{x}^2=1$
Implies that $\text{x}=1,-1$
But $\text{x}>0$
Implies that $\text{x}=1$ 
Now, $\text{f}''(\text{x})=\frac{1}{\text{x}^{3}}$
At x = 1
$\text{f}''(1)=\frac{2}{(1)^{3}}1=2>0$
Therefore, x = 1 is a point minimum.
Thus, the local minimum value is given by 
$\text{f}(1)=1+\frac{1}{1}=1+1=2$

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