Question

The function $f\left(x\right)=\frac{x}{2}+\frac{2}{x}$ has a local minimum at

• A. $x=-2$
• B. $x=0$
• C. $x=1$
• D. $x=2$

Answer 1

Answer: (A), (D)

The correct options are
A $x=-2$
D $x=2$

first find the first derivative of function

$=\frac{d}{dx}(\frac{x}{2}+\frac{2}{x})$

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

now equate the $f^1\left(x\right)$ to zero we get the local minimum values

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

$\frac{x^2-4}{2x^2}=0$

$x^2=4$

$x=2,-2$