# 5.4 Monotonic functions  (Page 3/3)

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$\text{Range}=\left[f\left({x}_{1}\right),f\left({x}_{2}\right)\right]$

We shall study this aspect of finding range in detail in a separate module.

## Non-decreasing function or increasing

The successive value of function increases or remains constant as the value of the independent variable increases. In other words, the preceding values are less than or equal to successive values that follow. Mathematically,

$\text{If}\phantom{\rule{1em}{0ex}}{x}_{1}<{x}_{2}\phantom{\rule{1em}{0ex}}\text{then}\phantom{\rule{1em}{0ex}}f\left({x}_{1}\right)\le f\left({x}_{2}\right)$

As f( ${x}_{1}$ )≤f( ${x}_{2}$ ) for all ${x}_{1}$ , ${x}_{2}$ ∈X, the difference “f(x+h) – f(x)” is non-negative for “h”, however small. This implies that the first derivative of function is non-negative. If we think of possibility, then we can realize that tangent to the function curve can be parallel to x-axis for a subset of X, while curve is increasing overall in the interval. It means that first derivative can be equal to zero points or sub-intervals in which it is increasing. Thus, for non-decreasing function,

$f\prime \left(x\right)\ge 0;\phantom{\rule{1em}{0ex}}\text{Equality sign holds for few points or a continuous section in X}$

For increasing function, if ${x}_{1}$ < ${x}_{2}$ , then f( ${x}_{1}$ ) ≤ f( ${x}_{2}$ ), for all ${x}_{1}$ , ${x}_{2}$ ∈X. This means that there may be same function values for different values of x. This is “many one” relation and as such function is not invertible in X.

## Strictly decreasing function

The successive value of function decreases as the value of the independent variable increases. In other words, the preceding values are greater than successive values that follow. Mathematically,

$\text{If}\phantom{\rule{1em}{0ex}}{x}_{1}<{x}_{2}\phantom{\rule{1em}{0ex}}\text{then}\phantom{\rule{1em}{0ex}}f\left({x}_{1}\right)>f\left({x}_{2}\right)$

Problem : Determine monotonic nature of the function in the interval (-∞,0].

$y={x}^{2}$

Solution : Let ${x}_{1}$ and ${x}_{2}$ belong to the interval [0,∞) such that ${x}_{1}$ < ${x}_{2}$ . Multiplying inequality with ${x}_{1}$ (a negative number) changes the nature of inequality :

$⇒{x}_{1}^{2}>{x}_{1}{x}_{2}$

Multiplying inequality with ${x}_{2}$ (a negative number) changes the nature of inequality :

$⇒{x}_{1}{x}_{2}>{x}_{2}^{2}$

Combining two inequalities,

$⇒{x}_{1}^{2}>{x}_{2}^{2}$ $⇒f\left({x}_{1}\right)>f\left({x}_{2}\right)$

Thus, given function is strictly decreasing in (-∞,0].

As f( ${x}_{1}$ )>f( ${x}_{2}$ ) for all ${x}_{1}$ , ${x}_{2}$ ∈X, the difference “f(x+h) – f(x)” is negative for “h”, however small. This implies that the first derivative of function is negative. If we think of possibility, then we can realize that tangent to the function curve can be parallel to x-axis for couple of x values, while curve is continuously decreasing in the interval. It means that first derivative can be equal to zero for few points in the interval in which it is strictly decreasing. Thus, for strictly decreasing function,

$f\prime \left(x\right)\le 0;\phantom{\rule{1em}{0ex}}\text{Equality sign holds for points only - not on a continuous section in X}$

For strictly decreasing function, if ${x}_{1}$ < ${x}_{2}$ , then f( ${x}_{1}$ )>f( ${x}_{2}$ ), for all ${x}_{1}$ , ${x}_{2}$ ∈X. It means that all distinct x values correspond to distinct y values and vice-versa. Therefore, strictly decreasing function is one-one function i.e. a bijection and hence “invertible”. In other words, if a function has strict decreasing order, then it is invertible. Mathematically, we say that if f’(x) ≤ 0 (equality holding for points only); x∈X, then function is invertible in X. For example, consider sine function,

$f\left(x\right)=\mathrm{sin}x$ $⇒f\prime \left(x\right)=\mathrm{cos}x$

We know that cosx is negative in the interval [π/2, 3π/2]. Hence sine function is a strictly decreasing function in [π/2, 3π/2]and is invertible. Recall though that inverse sine function is not defined in this interval, but in basic interval about origin [-π/2,π/2].

The order of a function provides an easy technique to determine range of a continuous function, corresponding to a given domain interval. For example, if domain of a continuously decreasing function, f(x), is [ ${x}_{1}$ , ${x}_{2}$ ], then the least value of the function is f( ${x}_{2}$ ) and greatest value of the function is f( ${x}_{1}$ ). Hence, range of the function is :

$\text{Range}=\left[f\left({x}_{2}\right),f\left({x}_{1}\right)\right]$

We shall study this aspect of finding range in detail in a separate module.

## Non-increasing function or decreasing

The successive value of function decreases or remains constant as the value of the independent variable increases. In other words, the preceding values are greater than or equal to successive values that follow. Mathematically,

$\text{If}\phantom{\rule{1em}{0ex}}{x}_{1}<{x}_{2}\phantom{\rule{1em}{0ex}}\text{then}\phantom{\rule{1em}{0ex}}f\left({x}_{1}\right)\ge f\left({x}_{2}\right)$

As f( ${x}_{1}$ )≥ f( ${x}_{2}$ ) for all ${x}_{1}$ , ${x}_{2}$ ∈X, the difference “f(x+h) – f(x)” is non-positive for “h”, however small. This implies that the first derivative of function is non-positive. If we think of possibility, then we can realize that tangent to the function curve can be parallel to x-axis for a subset of X, while curve is decreasing overall in the interval. It means that first derivative can be equal to zero at points or in sub-intervals in which it is decreasing. Thus, for non-decreasing function,

$f\prime \left(x\right)\le 0;\phantom{\rule{1em}{0ex}}\text{Equality sign holds for few points or a continuous section in X}$

For decreasing function, if ${x}_{1}$ < ${x}_{2}$ , then f( ${x}_{1}$ ) ≥ f( ${x}_{2}$ ), for all ${x}_{1}$ , ${x}_{2}$ ∈X. This means that there may be same function values for different values of x. This is “many one” relation and as such function is not invertible in X.

#### Questions & Answers

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Professor
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Period of sin^6 3x+ cos^6 3x
Period of sin^6 3x+ cos^6 3x