# 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?

• What is $f(x)$? $y=f(x)$? – npisinp Apr 11 '14 at 2:32
• @npisinp Did that help? – Joshhw Apr 11 '14 at 2:33
• The derivative of $\dfrac{1}{x^2}$ is $\dfrac{-2}{x^3}$ not $\dfrac{-1}{x^2}$ – npisinp Apr 11 '14 at 2:34
• @npisinp sorry, I mis wrote it. the original function is now correct. – Joshhw Apr 11 '14 at 2:35

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)?