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$f(x)=\frac 1{x}, x\geq 1$

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?

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  • $\begingroup$ What is $f(x)$? $y=f(x)$? $\endgroup$
    – npisinp
    Apr 11, 2014 at 2:32
  • $\begingroup$ @npisinp Did that help? $\endgroup$
    – Joshhw
    Apr 11, 2014 at 2:33
  • $\begingroup$ The derivative of $\dfrac{1}{x^2}$ is $\dfrac{-2}{x^3}$ not $\dfrac{-1}{x^2}$ $\endgroup$
    – npisinp
    Apr 11, 2014 at 2:34
  • $\begingroup$ @npisinp sorry, I mis wrote it. the original function is now correct. $\endgroup$
    – Joshhw
    Apr 11, 2014 at 2:35

1 Answer 1

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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)?

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