Question about finding where the function increases and decreases on f ( x ) = 1 x . f ( x ) = 1 x , x ≥ 1I have been staring at this equation for a bit. Things I'm confused on.the derivative of this is: f ′ ( x ) = − 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 − 1 x 2 &gt; 0 cross multiply to make it: − 1 &gt; 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 &gt; or &lt; or = and solving for the x give a critical point?

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Question about finding where the function increases and decreases on   f  (  x  )  =      1    x  .  f  (  x  )  =      1    x    ,  x  ≥  1I have been staring at this equation for a bit. Things I&#039;m confused on.the derivative of this is:       f    ′    (  x  )  =            −      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             −      1              x      2        &amp;gt;  0 cross multiply to make it:   −  1  &amp;gt;      x    2   and if so once I square this does it make the result   x  =  −  1  ,  x  =  1? I&#039;m really lost here and it seems like it should be easier.Does setting the derivative to  &amp;gt; or &amp;lt; or = and solving for the x give a critical point?

Rene Jordan · Accepted Answer

Explanation: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   −      1          x      2       using software. Where is the function positive, where is it negative, and where are its roots (if it has any)?