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Sunday, 3 August 2014

CHAPTER 10 - Introduction to Monotonicity

Consider a function represented by the following graph:
For two different input arguments ${x_1}$ and ${x_2}$, where ${x_1} < {x_2},{y_1} = f\left( {{x_1}} \right)$ will always be less than ${y_2} = f\left( {{x_2}} \right)$.
That is,
 ${x_1} < {x_2}$ implies $f\left( {{x_1}} \right) < f\left( {{x_2}} \right)$
Such a function is called a strictly increasing function or a monotonically increasing function (The word ‘monotonically’ apparently has its origin in the word monotonous; for example, a monotonous routine is one in which one follows the same routine repeatedly or continuously; similarly a monotonically increasing function is one that increases continuously).
Now, consider $f\left( x \right) = \left[ x \right]$. For this function
 ${x_1} < {x_2}$ does not always imply $f\left( {{x_1}} \right) < f\left( {{x_2}} \right)$
However,
 ${x_1} < {x_2}$ does imply $f\left( {{x_1}} \right) \le f\left( {{x_2}} \right)$
In other words, $f\left( x \right) = \left[ x \right]$ is not strictly (or monotonically) increasing. It will nevertheless be termed increasing.
Now consider a function represented by the following graph:
For two different input arguments ${x_1}$ and ${x_2}$, where ${x_1} < {x_2},{y_1} = f\left( {{x_1}} \right)$ will always be greater than ${y_2} = f\left( {{x_2}} \right)$.
That is,
 ${x_1} < {x_2} \Rightarrow f\left( {{x_1}} \right) > f\left( {{x_2}} \right)$
Such a function is called a strictly decreasing function or a monotonically decreasing function.
Now consider $f\left( x \right) = - \left[ x \right]$. For this function
 ${x_1} < {x_2}$ does not imply $f\left( {{x_1}} \right) > f\left( {{x_2}} \right)$
However,
 ${x_1} < {x_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f\left( {{x_1}} \right) \ge f\left( {{x_2}} \right)$
Therefore, $f\left( x \right) = - \left[ x \right]$ is not strictly decreasing. It would only be termed decreasing.
The following table lists down a few examples of functions and their behaviour in different intervals. You are urged to verify all the assertions listed on your own.
 Function Behaviour $f\left( x \right) = x$ Strictly increasing on $\mathbb{R}$ $f\left( x \right) = {x^2}$ Strictly decreasing on $\left( { - \infty ,0} \right]$ Strictly increasing on $\left[ {0,\infty } \right)$ $f\left( x \right) = \sqrt x$ Strictly increasing on $\left[ {0,\infty } \right)$ $f\left( x \right) = {x^3}$ Strictly increasing on $\mathbb{R}$ $f\left( x \right) = \left| x \right|$ Strictly decreasing on $\left( { - \infty ,0} \right]$ Strictly increasing on $\left[ {0,\infty } \right)$ $f\left( x \right) = \dfrac{1}{x}$ Neither increasing nor decreasing on $\mathbb{R}$. Strictly decreasing on $\left( { - \infty ,0} \right)$ Strictly decreasing on $\left( {0,\infty } \right)$ $f\left( x \right) = \left[ x \right]$ Increasing on $\mathbb{R}$ $f\left( x \right) = \left\{ x \right\}$ Neither increasing nor decreasing on $\mathbb{R}$ However, strictly increasing on $\left[ {n,n + 1} \right)$ where $n \in\mathbb{Z}$ $f\left( x \right) = \sin x$ Neither increasing nor decreasing on $\mathbb{R}$. Strictly increasing on $[(\,2n - \dfrac{1}{2}\,)\pi ,(2n + \dfrac{1}{2})\pi ];n \in\mathbb{Z}$ Strictly decreasing on $[(2n + \dfrac{1}{2})\pi ,(2n + \dfrac{3}{2})\pi ];n \in\mathbb{Z}$ $f\left( x \right) = \cos x$ Neither increasing nor decreasing on $\mathbb{R}$. Strictly increasing on $[(2n - 1)\pi ,2n\pi ];n \in\mathbb{Z}$ Strictly decreasing on $[2n\pi ,(2n + 1)\pi ];n \in\mathbb{Z}$ $f\left( x \right) = \tan x$ Neither increasing nor decreasing on $\mathbb{R}$ .Strictly increasing on $\left( {\left( {n - \dfrac{1}{2}} \right)\pi ,\left( {n + \dfrac{1}{2}} \right)\pi } \right);n \in\mathbb{Z}$ $f\left( x \right) = {e^x}$ Strictly increasing on $\mathbb{R}$ $f\left( x \right) = {e^{ - x}}$ Strictly decreasing on $\mathbb{R}$ $f\left( x \right) = \ln x$ Strictly increasing on $\left( {0,\infty } \right)$