Monotonicity and Extremum of Functions

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 Functions – Concept and Overview

The monotonicity of composite functions such as $f(g(x))$ and $g(f(x))$ depends on whether the individual functions $f(x)$ and $g(x)$ are increasing or decreasing. Understanding this helps in analyzing how a composite function behaves — whether it rises, falls, or remains constant - which is an essential concept in calculus and graph analysis.

Relationship Between $f(x)$ and $g(x)$

If $f(x)$ is an increasing function and $g(x)$ is a decreasing function, then for $x_2 > x_1$, we have

$f(x_2) \geq f(x_1)$ and $g(x_2) \leq g(x_1)$

Therefore, for $x_2 > x_1$,

$f(g(x_2)) \leq f(g(x_1))$ and $g(f(x_2)) \leq g(f(x_1))$

This means both $f(g(x))$ and $g(f(x))$ are decreasing functions.

If both $f(x)$ and $g(x)$ are either increasing or both are decreasing, then the composite functions $f(g(x))$ and $g(f(x))$ are increasing as well.

Monotonicity of Composite Functions Using Derivatives

When both $f(x)$ and $g(x)$ are differentiable, the derivative test gives a clear way to check monotonicity.

Case 1 – $f(x)$ Increasing and $g(x)$ Decreasing

If $f(x)$ and $g(x)$ are differentiable, with $f(x)$ increasing and $g(x)$ decreasing, then

$f'(x) \geq 0$ and $g'(x) \leq 0$

Now, the derivative of the composite function is

$(f(g(x)))' = f'(g(x)) \cdot g'(x)$

Since $f'(g(x)) \geq 0$ and $g'(x) \leq 0$,

$(f(g(x)))' \leq 0$

Therefore, $f(g(x))$ is a decreasing function.

Case 2 – Other Combinations of Monotonic Functions

The monotonic nature of the composite function depends on the signs of the derivatives $f'(x)$ and $g'(x)$. The possible cases are summarized below:

Here, (+) indicates strictly increasing and (−) indicates strictly decreasing.

Key Notes

• The monotonicity of composite functions $f(g(x))$ and $g(f(x))$ depends on whether $f(x)$ and $g(x)$ are increasing or decreasing.

• The derivative signs $f'(x)$ and $g'(x)$ determine the nature of the composite function.

• If both are increasing or both decreasing, the composite is increasing.

• If one is increasing and the other decreasing, the composite function is 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 conditions below 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.

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)

List of topics related to Monotonicity and Extremum of Functions

This section lists the key concepts connected to monotonic behavior and extrema of functions, including icontinuity and differentiability and continuity of composite functions, etc.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Continuity of Composite Function

Continuity And Differentiability

NCERT Resources

A compilation of NCERT-recommended learning materials from Class 12 Maths Chapter 5, designed to strengthen your understanding of continuity, differentiability, and monotonicity.

NCERT Class 12 Maths Notes for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Solutions for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Exemplar Solutions for Chapter 5 - Continuity and Differentiability

Practice Questions based on Monotonicity and Extremum of Functions

This section offers exam-based MCQs and practice problems designed to test your grasp of monotonic functions, stationary points, and extrema using $f'(x)$ and $f''(x)$.

Monotonicity And Extremum Of Functions- Practice Question MCQ

We have shared below the links to practice questions on the related topics to Monotonicity and Extremum of Functions