Monotonicity and Extremum of Functions

Monotonicity is an important concept in calculus. It is useful in understanding the relationship between curves and their slopes. The monotonic function is either increasing or decreasing. These concepts of monotonicity have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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This Story also Contains

1. Concavity
2. Monotonicity (Increasing and Decreasing Function)
3. Non-Monotonic Function and Critical Point
4. Solved Examples Based on Monotonicity
5. Summary

In this article, we will cover the concept of Monotonicity. This topic falls under the broader category of Calculus, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of twenty-one questions have been asked on this topic in JEE Main from 2013 to 2023, including one question in 2013, one in 2016, one in 2019, one in 2020, six questions in 2021, eight in 2022, and three in 2023.

Concavity

If $f^{\prime \prime}(x)>0$ in the interval $(a, b)$ then shape of $\mathrm{f}(\mathrm{x})$ in interval $(a, b)$ is concave when observed from upwards or convex down.

For convexity:
If $f^{\prime \prime}(x)<0$ in the interval $(a, b)$ then it is convex upward or concave down.

In case of graphs,

When you draw a tangent at any point on the curve, if the entire curve lies above the tangent, in this case, the curve is called a concave upward curve.

And if the entire curve lies below the tangent then the curve is called a concave downward curve.

Monotonicity (Increasing and Decreasing Function)

A function is said to be monotonic if it is either increasing or decreasing in its entire domain. By a monotonic function f in an interval I, we mean that f is either increasing in the Given domain or decreasing in a given domain.

Increasing Function

A function $f(x)$ is increasing in $[a, b]$ if $f\left(x_2\right) \geq f\left(x_1\right)$ for all $x_2>x_1$, where $x_1, x_2 \in[a, b]$.
If a function is differentiable, then $\frac{d}{d x}(f(x)) \geq 0 \quad \forall x \in(a, b)$
A function is said to be increasing if it is increasing in its entire domain.
Example:

• $f(x)=x$ is increasing in $R$. (As $f^{\prime}(x)=1$, so $f^{\prime}(x) \geq 0$ for all values of $x$ in $R$, so it is an increasing in $R$ ).
• $f(x)=\tan ^{-1} x$, is also an increasing function on $R$ as $f^{\prime}(x) \geq 0$ for all real values of $x$.

• - $f(x)=[x]$ is also an increasing function on $R$. Its differentiation is not defined at all points, but from its graph we can see that on giving higher value of $x$ to this function, it returns equal or higher value of $y$. For this function $x_2>$ $x_1$ implies $f\left(x_2\right) \geq f\left(x_1\right)$. Hence it is an increasing function.

Note:
These functions are also simply called 'increasing functions' as they are increasing in their entire domains.
$f(x)=\ln (x)$ is increasing function as it is increasing in its entire domain but it is not increasing in $R$ (as it is not defined for $x<0$ and $x=0$ )

So tangent to the curve, $f(x)$ at each point makes an acute angle with a positive direction of $x$-axis or parallel to the $x$-axis.

A function $y=f(x)$ is called an increasing function in an interval $I$.
for $x_1<x_2 \Rightarrow f\left(x_1\right) \leq f\left(x_2\right)$
or for $x_1>x_2 \Rightarrow f\left(x_1\right) \geq f\left(x_2\right)$
Condition for increasing functions
Where $f(x)$ is continuous and differentiable for $(a, b)$
For increasing function tangents drawn at any point on it make an acute slope with a positive $x$-axis.

$
\begin{aligned}
& M_T=\tan \theta \geq 0 \\
& \therefore \quad \frac{d y}{d x}=f^{\prime}(x) \geq 0 \text { for } x \in(a, b)
\end{aligned}
$

Strictly Increasing Function

A function $f(x)$ is strictly increasing in interval $[a, b]$ if $f\left(x_2\right)>f\left(x_1\right)$ for all $x_2>x_1$, where $x_1$, $x_2 \in[a, b]$.

If a function is differentiable, then

$
\frac{d}{d x}(f(x))>0 \quad \forall x \in(a, b)
$

So tangent to the curve, $f(x)$ at each point makes an acute angle with the positive direction of the $x$-axis.

Example: $f(x)=x$ is strictly increasing but $f(x)=[x]$ is not strictly increasing
Note:
If $f^{\prime}(x)=0$ at some discrete points (if number of such points can be counted), and at other points $f^{\prime}(x)>0$, still the function is strictly increasing function.

Example
Consider $f(x)=[x]$, where $[$.$]$ is the greatest integer function.
For this function $x_2>x_1$ does not always implies $f\left(x_2\right)>f\left(x_1\right)$
However, $x_2>x_1$ does imply $f\left(x_2\right) \geq f\left(x_1\right)$
So, $f(x)=[x]$ is increasing function but not a strictly increasing function.

Let's look into some more examples,
Functions $\mathrm{e}^{\mathrm{x}}, \mathrm{a}^{\mathrm{x}}(a>1), \mathrm{x}^3+\mathrm{x}$ are strictly increasing functions in their entire domain.

$
\frac{d}{d x}\left(e^x\right)=e^x>0 \text { and } \frac{d}{d x}\left(x^3+x\right)=3 x^2+1>0, \quad \forall x
$

Strictly Increasing functions can be classified as:

1. Concave up: When $f’(x) > 0$ and $f''(x) > 0 \quad∀\quad x ∈$ domain

2. Concave down: When $f’(x) > 0$ and $f''(x) < 0 \quad∀\quad x ∈$ domain

3. When $f’(x) > 0$ and $f”''(x) = 0 \quad∀ \quad x ∈$ domain

Decreasing Function

A function $f(x)$ is decreasing in the interval $[a, b]$ if $f\left(x_2\right) \leq f\left(x_1\right)$ for all $x_2>x_1$, where $x_1, x_2 \in[a, b]$
If a function is differentiable, then $\frac{d}{d x}(f(x)) \leq 0 \quad \forall x \in(a, b)$
Example
- $f(x)=-x$ is decreasing in $R\left(\right.$ As $f^{\prime}(x)=-1$, so $f^{\prime}(x)<0$ for all real values of $x$. We can also see that it is decreasing from its graph)
- $f(x)=e^{-x}$ is decreasing in $R$ (As $f^{\prime}(x)=-e^{-x}$, so $f^{\prime}(x)<0$ for all real values of $x$. We can also see that it is decreasing from its graph)
- $f(x)=\cot (x)$ is decreasing in $(0, \pi)$
- $f(x)=\cot ^{-1}(x)$ is decreasing in $R$

A function is said to be decreasing if it is decreasing in its entire domain.
So tangent to the curve, $\mathrm{f}(\mathrm{x})$ at each point makes an obtuse angle with the positive direction of $x$-axis or parallel to the $x$-axis.

Strictly Decreasing Function

A function $f(x)$ is strictly decreasing in its domain ($Df$) if $f\left(x_2\right)<f\left(x_1\right)$ for all $x_2>x_1$, where $\mathrm{x}_1, \mathrm{x}_2 \in$ $Df$.If a function is differentiable in domain ($Df$) then

$
\frac{d}{d x}(f(x))<0 \quad \forall x
$

So tangent to the curve, $f(x)$ at each point makes an obtuse angle with the positive direction of the $x$-axis.

For example, functions $\mathrm{e}^{-\mathrm{x}}$ and $-\mathrm{x}^3$ are strictly decreasing functions.
Note:
If $f^{\prime}(x)=0$ at some discrete points (if a number of such points can be counted), and at other points $f^{\prime}(x)>0$, still the function is strictly increasing function.

NOTE:
- If a function is not differentiable at all points, this does not mean that the function is not increasing or decreasing. A function may increase or decrease on an interval without having a derivative defined at all points.

For example, $y=x^{1 / 3}$ is increasing everywhere including $x=0$, but the derivative is not defined at this point as the function has vertical tangent.

Decreasing functions can be classified as:

1. Concave up: When $f’(x) < 0$ and $f''(x) > 0 \quad∀\quad x ∈$ domain

2. Concave down: When $f’(x) < 0 and f''(x) < 0\quad ∀\quad x ∈$ domain

3. When $f’(x) > 0 and f''(x) = 0 \quad∀\quad x ∈$domain

Monotonicity of Composite Function

The nature of monotonicity of composite functions $f(g(x))$ and $g(f(x))$ depends on the nature of the function $f(x)$ and $g(x)$.

If $f(x)$ is increasing function and $g(x)$ is decreasing function, then for $x_2>x_1$, we have $f\left(x_2\right) \geq f\left(x_1\right)$ and $g\left(x_2\right) \leq g\left(x_1\right)$.

So, for $x_2>x_1$, we have $f\left(g\left(x_2\right)\right) \leq f\left(g\left(x_1\right)\right)$ and $g\left(f\left(x_2\right)\right) \leq g\left(f\left(x_1\right)\right)$.
Thus, $f(g(x))$ is a decreasing function and also, $g(f(x))$ is also a decreasing function.
If both $f(x)$ and $g(x)$ are increasing or decreasing functions, then $f(g(x))$ and $g(f(x))$, i.e., both composite functions are increasing.

For differentiable functions, we can prove it in another way
If $f(x)$ and $g(x)$ are differentiable function, with $f(x)$ increasing and $g(x)$ decreasing, then

$
\begin{aligned}
& \quad \mathrm{f}^{\prime}(\mathrm{x}) \geq 0 \quad \text { and } \mathrm{g}^{\prime}(\mathrm{x}) \leq 0 \\
& \therefore \quad(\mathrm{f}(\mathrm{~g}(\mathrm{x})))^{\prime}=\mathrm{f}^{\prime}(\mathrm{g}(\mathrm{x})) \mathrm{g}^{\prime}(\mathrm{x}) \leq 0 \quad\left[\text { as } \mathrm{f}^{\prime}(\mathrm{g}(\mathrm{x})) \leq 0\right]
\end{aligned}
$

$\therefore \quad \mathrm{f}(\mathrm{g}(\mathrm{x}))$ is a decreasing function
Similarly, all the possibilities of the nature of the composite function $f(g(x))$ and $g(f(x))$ are given below

AID TO MEMORY:

Where (+) means strictly increasing and (-) means strictly decreasing.

Non-Monotonic Function and Critical Point

A function that is neither always increasing nor always decreasing in its domain is called non-monotonic function.

For example,

$f(x) = \sin x$, which is increasing in the first quadrant and the fourth quadrant and decreasing in the second and third quadrants.

Consider another function, $y = f(x) = |x² - 2| $

$f(x)$ is increases in $[-√2, 0]$ and $[√2, ∞ )$ and decreases in $(-∞, -√2]$ and $[0,√2]$

Hence this function is non-monotonic.

Critical Points

A critical point of a function is a point where its derivative does not exist or its derivative is equal to zero.

All the values of ' $x$ ' obtained by the below conditions are said to be the critical points.
1. $f(x)$ does not exists
2. $f^{\prime}(x)$ does not exists
3. $f^{\prime}(x)=0$

Critical points are interior points of the intervals.
For the function $f(x)=\left|x^2-4\right|$, critical points are $x=+2,-2$, and $x=0$ where its derivative is zero.

Recommended Video Based on Monotonicity

Solved Examples Based on Monotonicity

Example 1: If $m$ is the minimum value of $k$ for which the function $f(x)=x \sqrt{k x-x^2}$ is increasing in the interval $[0,3]$ and $M$ is the maximum value of $f_{\text {in }}[0,3]_{\text {when }} k=m$, then the ordered pair $(m, M)$ is equal to : [JEE Main 2019]
1) $(4,3 \sqrt{3})$
2) $(3,3 \sqrt{3})$
3) $(5,3 \sqrt{6})$
4) $(4,3 \sqrt{2})$

Solution
Condition for increasing functions -
For increasing function tangents drawn at any point on it makes an acute slope with positive $x$ axis.

$
\begin{aligned}
& M_T=\tan \theta \geq 0 \\
& \therefore \quad \frac{d y}{d x}=f^{\prime}(x) \geq 0 \text { for } x \epsilon(a, b)
\end{aligned}
$

- wherein

Where $f(x)$ is continuous and differentiable for $(a,b)$
Method for maxima or minima -
By second derivative method:
Step 1. find values of $x$ for $\frac{d y}{d x}=0$
Stcp 2. $x=x_0$ is a point of local maximum if $f^{\prime \prime}(x)<0$ and local minimum if $f^{\prime \prime}(x)>0$
- wherein

Where $y=f(x)$

$
\begin{aligned}
& \frac{d y}{d x}=f^{\prime}(x) \\
& f(x)=x \sqrt{k x-x^2} \\
& f^{\prime}(x)=3 k x-4 x^2 \cdot \frac{1}{2 \sqrt{k-x^2}}
\end{aligned}
$

$
\begin{aligned}
& \text { For } \uparrow f^{\prime}(x) \geqslant 0 \\
& k x-x^2 \geqslant 0 \\
& x^2-k x \leq 0 \\
& x(x-k) \leq 0 \\
& +v e \quad x \geqslant 3
\end{aligned}
$

$\& 3 k x-4 x^2 \geqslant 0$
$4 x^2-3 k x \leq 0$
$4 x\left(x-\frac{3 k}{4}\right) \leq 0$
$x-\frac{3 k}{4} \leq 0$
$3-\frac{3 k}{4} \leq 0$
$k \geq 4$
minimurn value of $k$ is $m=4$

$
\begin{aligned}
& \begin{aligned}
f(x) & =x \sqrt{k x-x^2} \\
& =3 \sqrt{4 \times 3-3^2} \\
& =3 \sqrt{3}, \quad M=3 \sqrt{3}
\end{aligned} \\
& (4,3 \sqrt{3})
\end{aligned}
$

minimum value of $x=3$

Example 2: Let $f: R \rightarrow R$ be defined as
$f(x)=\left\{\begin{array}{cc}-55 x & \text { if } x<-5 \\ 2 x^3-3 x^2-120 x & \text { if }-5 \leq x \leq 4 \\ 2 x^3-3 x^2-36 x-336, & \text { if } x>4,\end{array}\right.$ $A=\{x \in R: f$ is increasing $\}$. Then $A$ is equal to : [JEE Main 2021]
1) $(-5, \infty)$
2) $(-5,-4) \cup(4, \infty)$
3) $(-\infty,-5) \cup(4, \infty)$
4) $(-\infty,-5) \cup(-4, \infty)$

solution

$
\begin{aligned}
& f(x)=\left\{\begin{array}{cc}
-35 x & \text { if } x<-5 \\
2 x^3-3 x^2-120 x & \text { if }-5 \leq x \leq 4 \\
2 x^3-3 x^2-36 x-336, & \text { if } x>4
\end{array}\right. \\
& f^{\prime}(x)=\left\{\begin{array}{cc}
-55 ; & x<-5 \\
6(x-5)(x+4) ; & -5<x<4 \\
6(x-3)(x+2) ; & x>4
\end{array}\right.
\end{aligned}
$

$\mathrm{f}(\mathrm{x})$ is increasing in

$
x \in(-5,-4) \cup(4, \infty)
$

Example 3: Let $f(x)=\sin ^4 x+\cos ^4 x$. Then $f$ is an increasing function in the interval : [JEE Main $2016]$
$\begin{aligned}&
1) ] 0, \frac{\pi}{4}[ \\ &
2) ] \frac{\pi}{4}, \frac{\pi}{2}[ \\ &
3) ] \frac{\pi}{2}, \frac{5 \pi}{8}[ \\ &
4) ] \frac{5 \pi}{8}, \frac{3 \pi}{4}[ \end{aligned}$
solution

$
\begin{aligned}
& f(x)=\sin ^4 x+\cos ^4 x \\
& f^{\prime}(x)=4 \sin ^3 x \cos x-4 \cos ^3 x \sin x \\
& f^{\prime}(x)=4 \sin x \cos x\left(\sin ^2 x-\cos ^2 x\right) \\
& f^{\prime}(x)-2 \sin 2 x \cdot \cos 2 x \\
& f^{\prime}(x)-\sin 4 x>0 \\
& f^{\prime}(x)=\sin 4 x<0 \\
& \frac{\Rightarrow}{\pi} \pi<4 x<2 \pi \\
& \frac{\pi}{4}<x<\frac{\pi}{2}
\end{aligned}
$

Example 4: The number of distinet real roots of the equation $x^7-7 x-2=0$ is: [JEE Main $2022]$
1) $5$
2) $7$
3) $1$
4) $3$

Solution

$\begin{aligned} & x^7-7 x-2=0 \\ & \operatorname{lnt} f(x)=x^7-7 x-2 \\ & f^5(x)=7\left(x^5-1\right)=7\left(x^3-1\right)\left(x^3+1\right) \\ & =7(x-1)(x+1)\left(x^2+x+1\right)\left(x^2-x+1\right)\end{aligned}$

at $x=1 ; f(x)=1+7-2=-8 \quad x=-1 ; f(x)=-1+7-2=4$

Hence 3 distinct solutions
Example 5: The function $\mathrm{f}(\mathrm{x})=\mathrm{xe}^{\mathrm{x}(1-\mathrm{x})}, \mathrm{x} \in \mathbb{R}$, is: [JEE Main 2022]
1) increasing in $\left(-\frac{1}{2}, 1\right)$
2) decreasing in $\left(\frac{1}{2}, 2\right)$
3) increasing in $\left(-1,-\frac{1}{2}\right)$
4) decreasing in $\left(-\frac{1}{2}, \frac{1}{2}\right)$
solution

$
\begin{aligned}
f^{\prime}(x) & =e^{x(1-x)}+x e^{x(1-x)} \cdot(1-2 x) \\
& =e^{x(1-x)}\left[1+x-2 x^2\right] \\
& =-e^{x(1-x)}(2 x+1)(x-1)
\end{aligned}
$

$\therefore$ option (A)

Summary

Monotonicity is an important part of the mathematics. A function is either increasing or decreasing. A function is said to be monotonic if it is either increasing or decreasing in its entire domain. This concept is used in various fields of physics and chemistry.