The min of $x + k/x$ $f(x) = x + \dfrac{k}{x} $ ($k > 0$ and $x > 0$)
given that the k is a constant, how do I solve the x that makes f(x) minimum?

$$x+\frac{k}{x}=\frac{1}{x}\Big[(x-\sqrt k)^2+2\sqrt k x\Big]$$
$$=2\sqrt k+\frac{(x-\sqrt k)^2}{x}\geq2\sqrt k$$
Takes its minimum at $x=\sqrt k$
I have assumed $x>0$. Do the $x<0$ for yourself. Hint: This time the function has a maximum!
