Find the minimum of this function

Substituting $x = 1$ , then $f ( x ) = 1 + 1 = 2$ .

$\text{By AM-GM Inequality,}$ $x+\dfrac{1}{x} \geq 2\sqrt{x\times \dfrac{1}{x} }$ $x>0 \implies \sqrt{x\times \dfrac{1}{x} } =1$ $\implies \min\left(x+\dfrac{1}{x}\right) =2$

f(x) = ( x + (1/x)); $\frac{df(x)}{dx}$ = 1 - $\frac{1}{x*x}$ ; Min occurs when f '(x) = 0 and when f ''(x) is positive. Calculating f '(x) ( 1 -( 1/x*x)) = 0 occurs at either x = 1 or x = -1. f ''(x) = 2/x^3. f ''(1) = 2/3; f ''(-1) = (-2/3). Hence we can conclude that minimum occurs at x = 1 , follows that f(1) = 2;
The minimum value of f(x) = 2.