Monotonicity and Extremum of Functions - Practice Questions & MCQ

Monotonicity (Increasing and Decreasing Function)

A function is said to be monotonic if it is either increasing or decreasing in its entire 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 $x_1, x_2 \in[a, b]$.

$
\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 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 $x$. $f\left(x_1\right)$. 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 an 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 $the x$axis or parallel to the $x$axis.

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 $\mathrm{f}^{\prime}(\mathrm{x})=0$ at some discrete points (if several such points can be counted), and at other points $\mathrm{f}^{\prime}(\mathrm{x})>0$, 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 an increasing function but not a strictly increasing function.

Let's look into some more examples,
Functions $\mathrm{e}^{\mathrm{x}}, \mathrm{a}^{\mathrm{x}}(\mathrm{a}>1), \mathrm{x}^3+\mathrm{x}$ 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
$

Concavity

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.

Strictly Increasing functions can be classified as:

Concave up: When $\mathrm{f}^{\prime}(\mathrm{x})>0$ and $\mathrm{f}^{\prime}(\mathrm{x})>0 \forall \mathrm{x} \in$ domain

Concave down: When$f^{\prime}(x)>0$ and $f^{\prime \prime}(x)<0 \forall x \in$ domain

When $f^{\prime}(x)>0$ and $f^{\prime}(x)=0 \quad \forall x \in$ 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 $R\left(A s f^{\prime}(x)=-1\right.$, 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, $f(x)$ 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 $x_1, x_2 \in D f$ If a function is differentiable in domain (Df) then.

$
\frac{d}{d x}(f(x))<0 \quad \forall x .
$
So tangent to the curve, $\mathrm{f}(\mathrm{x})$ 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 several 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 a vertical tangent.

Decreasing functions can be classified as:

Concave up: When $f^{\prime}(x)<0$ and $f^{\prime}(x)>0 \forall x \in$ domain

Concave down:When $f^{\prime}(x)<0$ and $f^{\prime \prime}(x)<0 \forall x \in$ domain

When $f^{\prime}(x)>0$ and $f^{\prime \prime}(x)=0 \quad \forall x \in$ domain

Monotonicity of Composite Function

The nature of the monotonicity of composite functions $f(\mathrm{~g}(\mathrm{x}))$ and $\mathrm{g}(\mathrm{f}(\mathrm{x}))$ depends on the nature of the functions $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$.

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, $\mathrm{f}(\mathrm{g}(\mathrm{x}))$ is a decreasing function and also, and $\mathrm{g}(\mathrm{f}(\mathrm{x}))$ 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}
& \mathrm{f}^{\prime}(\mathrm{x}) \geq 0 \quad \text { and } \quad \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 which is neither always increasing nor always decreasing in its domain is called a non-monotonic function.

For example,

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

Consider another function, $y=f(x)=\left|x^2-2\right|$

$f(x)$ is increases in $[-\sqrt{ } 2,0]$ and $[\sqrt{ } 2, \infty)$ and decreases in $(-\infty,-\sqrt{2}]$ and $[0, \sqrt{2}]$
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. $\mathrm{f}(\mathrm{x})$ 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.