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The function f(x) = (1\x) is monotonic on its domain ( true or false) ?
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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 - .
The denominator 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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After reading the post by @math_tutor2020
https://www.algebra.com/algebra/homework/complex/Complex_Numbers.faq.question.1210389.html
This problem is an elementary .
As I explained, locally and piece-wisely, the function is monotonically decreasing.
But globally, it is NOT monotonic.
It is OBVIOUS: compare these values
x -2 -1 1 2
1/x -0.5 -1 1 0.5
decease decrease
increase
They show non-monotonic behavior.
Tutor @math_tutor2000 fell into this trap.