Minimum value of $f(x) = {x^2 +18 \over x}$

Question

How can I calculate the minimum value of

$$f(x) = {x^2 +18\over x}$$

without using a graphing calculator or any computer software ?

Answer 1

if $x>0$ then we get by $AM-GM$ $$x+\frac{18}{x}\geq 2\sqrt{18}=6\sqrt{2}$$

Answer 2

Rewrite the function as \begin{eqnarray*} x+ \frac{18}{x}= \frac{1}{x} \underbrace{(x-\sqrt{18})^2}_{ \text{min occurs at } x=\sqrt{18}}+ \underbrace{ 6 \sqrt{2} }_{\text{mininimum value of the function is } 6\sqrt{2} }. \end{eqnarray*}