Answer
$(\frac{\sqrt 2}{2},0)$ and $(-\frac{\sqrt 2}{2},0)$
Work Step by Step
Step 1. Find the derivative of the function $y=x-\frac{1}{2x}$ as $y'=1+\frac{1}{2x^2}$
Step 2. Let $y'=2$, we have $1+\frac{1}{2x^2}=2$ and $\frac{1}{x^2}=2$ which gives $x^2=\frac{1}{2}$ and $x=\pm\frac{\sqrt 2}{2}$
Step 3. We have $x_1=\frac{\sqrt 2}{2}$ and $x_2=-\frac{\sqrt 2}{2}$ corresponding to $y_1=\frac{\sqrt 2}{2}-\frac{1}{2\frac{\sqrt 2}{2}}=\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}=0$ and $y_2=-\frac{\sqrt 2}{2}+\frac{1}{2\frac{\sqrt 2}{2}}=-\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}=0$
Step 4. Thus the answer is yes and the points are $(\frac{\sqrt 2}{2},0)$ and $(-\frac{\sqrt 2}{2},0)$
