JEE Main · 2006 · 30 Apr 2006 · Offline · Application of Derivatives

Differentiate $f(x)=\frac{x}{2}+\frac{2}{x}$ to get $f^{\prime }(x)=\frac{1}{2}-\frac{2}{x^2}$, which vanishes when $x^2=4$, i.e. $x=\pm 2$. The second derivative is $f^{\prime \prime }(x)=\frac{4}{x^3}$. At $x=2$ this is $\frac{1}{2}>0$, marking a local minimum, whereas at $x=-2$ it is $-\frac{1}{2}<0$, a local maximum. The local minimum therefore occurs at $x=2$.