Find the equation of the normal at the point $(1,2)$ to the curve $y=x+\frac{1}{x}$. The answer in the textbook is listed as $x=1$, however I don't understand why this is.
My working:
$$y=x+\frac{1}{x}$$
$$f'(x)=1-x^{-2}$$
$$f'(x)=1-\frac{1}{x^{2}}$$
$$f'(1)=1-\frac{1}{1}=0$$
Therefore the gradient of the tangent to the curve at $x=1$ is $0$.
The gradient of the normal is given by: 
$$m_1m_2=-1$$
$$m_2=\frac{-1}{0}$$
As anything divided by $0$ is undefined, how do I then use this to find the the normal to the tangent, to the curve, at $x=1$? Is there a way of analytically working out tangents/normals when they are undefined, or is an alternative approach required?
EDIT: 
My alternative approach would entail ascertaining the tangent at $(1,2)$, which turns out to be $y=2$ - a horizontal line; which means that the normal would be perpendicular to this and is thus given by $x$-value at this point, i.e. $x=1$.  
Is a visual way the only method of producing an answer? Is there a better way of demonstrating this?

Perhaps a picture will clarify the situation.

