Question about finding where the function increases and decreases on $f(x)=\frac 1{x}$ $f(x)=\frac 1{x}, x\geq 1$
I have been staring at this equation for a bit. Things I'm confused on.
the derivative of this is: $f'(x)= \frac {-1}{x^2}$ now, how am I supposed to find where this derivative increases/decreases? Do I find the critical points first? by setting the derivative to 0? or do I solve it like $\frac {-1}{x^2} > 0$ cross multiply to make it: 
$-1>x^2$ and if so once I square this does it make the result x=-1, x= 1? I'm really lost here and it seems like it should be easier. 
does setting the derivative to > or < or = and solving for the x give a critical point?

Recall that intervals of increase and decrease of $f$ correspond to intervals of positivity and negativity of $f'$, and critical points of $f$ are where the roots of $f'$ are. Try graphing the function $-\frac{1}{x^2}$ using software. Where is the function positive, where is it negative, and where are its roots (if it has any)?
