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 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:

$\begin{array}{|c||c||c|}
\hline f^{\prime}(x) & g^{\prime}(x) & (f \circ g)^{\prime}(x) \text { and }(g \circ f)^{\prime}(x) \\
\hline \hline+ & + & + \\
\hline+ & - & - \\
\hline- & + & - \\
\hline- & - & + \\
\hline
\end{array}$

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.