Question 1210389
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        First of all,  I read this function as   f(x) = 1/x, 

        following to the standard rules commonly accepted in  Math.



The domain is the set of all real numbers except of 0 (zero).


Function   f(x) = 1/x   has two branches: one over the set of positive  'x',
another over the set of negative  'x'.


For each branch separately,  function   f(x) = 1/x   is monotonic in each  sub-domain.



Or, more accurately,  function   f(x) = 1/x   is monotonically decreases locally in each sib-domain.


To prove it,  take the derivative:  it is    - {{{1/x^2}}}.


The denominator  {{{x^2}}}  is always positive for all  'x'  of the domain;   so, the derivative is always negative.


You also can convince yourself by making a plot of this function.


For it,  go to website https:\\www.desmos.com/calculator/


Print the formula for the function   y = 1/x.


You will get the plot instantly.


The plot clearly shows that the function monotonically decreases as  'x'  increases in the domain.


So,  locally for each branch,  the function   f(x) = 1/x   monotonically decreases.
You have this answer proved formally and demonstrated / illustrated visually.



But we can not say that function   f(x) = 1/x   is globally decreasing:   when  'x'   moves 
from negative values to positive values,  the function  1/x  jumps from negative values to positive,
breaking monotonicity.


So,  locally this function is monotonically decreasing,  but globally it is not monotonic.



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            This problem is an elementary   {{{highlight(highlight(TRAP))}}}. 



As I explained in my post above,  locally and piece-wisely,  the function is monotonically decreasing.


But globally,  it is  NOT  monotonic.


It is  OBVIOUS:  compare these values of the function


<pre>
    x     -2     -1     1     2   

    1/x   -0.5   -1     1     0.5

          decrease      decrease
                 increase
</pre>

They show non-monotonic behavior.


Tutor &nbsp;@math_tutor2000 &nbsp;fell into this trap.