The function, $f(x) = x +\frac{a^2}{2x}, a > 0, x ≠ 0$ has a local maxima at

$x=\frac{a}{\sqrt(2)}$

$f(x)=x+\frac{a^2}{2x},\; a>0,\; x\ne0$

$\frac{dy}{dx}=1-\frac{a^2}{2x^2}$

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

$\frac{a^2}{2x^2}=1$

$x^2=\frac{a^2}{2}$

$x=\pm\frac{a}{\sqrt2}$

$\frac{d^2y}{dx^2}=\frac{a^2}{x^3}$

At $x=-\frac{a}{\sqrt2}$

$\frac{d^2y}{dx^2}<0$

Local maximum occurs at $x=-\frac{a}{\sqrt2}$