Continuity of Composite Function

Continuity is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which graphs of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of Continuity and Discontinuity have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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In this article, we will cover the concepts of Continuity of Composite Functions. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of nine questions have been asked on this concept, including two in 2013, one in 2014, one in 2019, one in 2021 and four in 2022.

Continuity of Composite Function

Function-

A relation from a set $A$ to a set $B$ is said to be a function from $A$ to $B$ if every element of set $A$ has one and only one image in set $B$.

OR
$A$ and $B$ are two non-empty sets, then a relation from $A$ to $B$ is said to be a function if each element $x$ in $A$ is assigned a unique element $f(x)$ in $B$, and it is written as
$f:A\to B$ and read as $f$ is a mapping from $A$ to $B$.

Function Function Not a function

Not a function

The third one is not a function because $d$ is not related(mapped) to any element in B.

Fourth is not a function as element a in A is mapped to more than one element in B .

Composition of function

Let $\mathrm{f}:\mathrm{A}\to \mathrm{B}$ and $\mathrm{g}:\mathrm{B}\to \mathrm{C}$ be two functions. Then the composition of f and $g$ is denoted by gof and defined as the function gof : $A\to C$ given by $g\circ f(x)=g(f(x))$

Symbol of Composition of Functions

The symbol of the composition of functions is $\circ$. It can also be shown without using this symbol but by using the brackets. i.e.,

• $(f\circ g)(x)=f(g(x))$ and is read as " $f$ of $g$ of $x$ ". Here, $g$ is the inner function and $f$ is the outer function.
• $(g\circ f)(x)=g(f(x))$ and is read as "$g$ of $f$ of $x$ ". Here, $f$ is the inner function and g is the outer function.

Theorem 2 Suppose $f$ and $g$ are real-valued functions such that $(f\circ g)$ is defined at c . If g is continuous at c and if $f$ is continuous at $\mathrm{g}(\mathrm{c})$, then ( $fo$ $\mathrm{g})$ is continuous at c . The following examples illustrate this theorem.

Example: Show that the function defined by $f(x)=\sin \left(x^2\right)$ is a continuous function.

Solution Observe that the function is defined for every real number. The function f may be thought of as a composition $\mathrm{g}\circ \mathrm{h}$ of the two functions $g$ and $h$ , where
$g(x)=\sin x$ and $h(x)=x^2$. Since both $g$ and $h$ are continuous functions, by Theorem 2, it can be deduced that $f$ is a continuous function.

Continuity:

If the function $f(x)$ is continuous at the point $x=a$ and the function $y=g(x)$ is continuous at the point $x=f(a)$, then the composite function $y=(g\circ f)(x)=g(f(x))$ is continuous at the point $x=a$.
Consider the function $f(x)=\frac{1}{1-x}$, which is discontinuous at $x=1$
If $g(x)=f(f(x))$
$g(x)$ will not be defined when $f(x)$ is not defined, so $g(x)$ is discontinuous at $x=1$
Also $g(x)=f(f(x))$ is discontinuous when $f(x)=1$
i.e. $\frac{1}{1-x}=1\Rightarrow x=0$

We ca check it by finding $g(x),g(x)=\frac{1}{1-f(x)}=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$
It is discontinuous at $x=0$
So, $g(x)=f(f(x))$ is discontinuous at $x=0$ and $x=1$

Now consider,

$\begin{array}{rl}\mathrm{h}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x})))& =\mathrm{f}\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)\\ & =\frac{1}{1-\frac{\mathrm{x}-1}{\mathrm{x}}}=\mathrm{x}\end{array}$

seems to be continuous, but it is discontinuous at $x=1$ and $x=0$ where $f(x)$ and $f(f(x))$ respectively are not defined.

Summary

The composition of functions involves using the output of one function as the input to another, forming a new function. This operation is associative and plays an important role in various mathematical and applied fields, including calculus, function transformation, and computer science. Understanding and working with function composition allows for the creation of complex functions from simpler building blocks.

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Solved Examples Based On the Continuity of Composite Functions:

Example 1: Let $f:[-1,3]\to R\text{be defined as}$ $f(x)=\{\begin{array}{rlrl}|x|+[x],& & -1& \le x<1\\ x+|x|,& & 1& \le x<2\\ x+[x],& & 2\le x\le 3\end{array}$ where [t] denotes the greatest integer less than or equal to $t$. Then $f$ is discontinuous at:
1) only one point
2) only two points
3) only three points
4) four or more points

Solution: Continuity of composite functions-

A composite function fog $(x)$ is continuous at $x=a$ if $g$ is continuous at $x=$ a and f is continuous at $\mathrm{g}(\mathrm{a})$.

$\begin{array}{rl}& f(x)=\{\begin{array}{rc}|x|+[x],& -1\le x<1\\ x+|x|,& 1\le x<2\\ x+[x],& 2\le x\le 3\end{array}\\ & f(x)=\{\begin{array}{cc}-x-1,& -1\le x<0\\ x+0,& 0\le x<1\\ 2x,& 1\le x<2\\ x+2,& 2\le x<3\\ x+3,& x=3\end{array}\end{array}$

$f(x)$ is discontinuous at $x=0,1,3$.
Hence, the answer is the option (3).

Example 2: If $f(x)=\{\begin{array}{l}\sin x,x\ne n\pi \\ 2,\text{otherwise}\end{array}$ and $g(x)=\{\begin{array}{l}n\in Z\\ x^2+1,x\ne 0,2\\ 4,x=0\\ 5,x=2\end{array}$ then the number of points in $[\frac{-\pi }{2},\frac{\pi }{2}]$ where $g\{f(x)\}\text{is discontinuous.}$
1) $1$
2) $0$
3) $2$
4) None of these

Solution:

$\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=\mathrm{n}\pi$ so check at $\mathrm{x}=0\underset{x\to 0^{+}}{lim}g\{f(x)\}=\underset{x\to 0^{+}}{lim}g\{\sin x\}=\underset{x\to 0^{+}}{lim}({\sin }^{2}x+1)=1$

similarly $\underset{x\to 0^{-}}{lim}g\{f(x)\}=1g\{f(0)\}=5\underset{x\to 0^{+}}{lim}g\{f(x)\}=\underset{x\to 0^{-}}{lim}g\{f(x)\}\ne g\{f(0)\}g\{f(x)\}$ is not continuous function at $\mathbf{x}=0$,

on any other values of $x$ it is continuous function in given interval

Hence, the answer is the option 1.

Example 3: Number of points where $f(g(x))$ is discontinuous if $f(x)=\frac{1}{x-6}$ and $g(x)=x^2+5$
1) $2$
2) $1$
3) $0$
4) $3$

Solution:

$g(x)=x^2+5$, which is always continuous
Now, $f(x)=\frac{1}{x-6}\Rightarrow f(g(x))=\frac{1}{g(x)-6}$

$f(g(x))=\frac{1}{(x^2+5)-6}=\frac{1}{x^2-1}$
Discontinuous at $\mathrm{x}=1,\mathrm{x}=-1$, as denominator $=0$ at these points
Hence, the answer is the option (1).

Example 4: If $f(x)=\frac{1}{x-6}$, then number of points where $f(f(x))$ is discontinuous is
1) $2$
2) $0$
3) $1$
4) $4$

Solution:

$f(x)=\frac{1}{x-6}$

$f(x)$ is not defined at $x=6$, so $g(x)$ is not defined and hence discontinuous at $x=6$

$\begin{array}{rl}& g(x)=f(f(x))=\frac{1}{f(x)-6}\\ & g(x)=f(f(x))=\frac{1}{\frac{1}{x-6}-6}\\ & g(x)=f(f(x))=\frac{x-6}{37-6x}\end{array}$

$\mathrm{f}(\mathrm{f}(\mathrm{x}))$ is not defined at $x=\frac{37}{6}$
So number of point where $f(f(x))$ is discontinuous is $2$

Hence, the answer is the option (1).

Example 5: The number of points where the function
$\mathrm{f}(\mathrm{x})=\{\begin{array}{ll}|2x^2-3x-7|& \text{if}x\le -1\\ [4x^2-1]& \text{if}-1<x<1\\ |x+1|+|x-2|& \text{if}x⩾1,[\mathrm{t}{]}_{\text{denotes the greatest integer}}⩽\mathrm{t}\text{, is discontinuous is}\end{array}$ $$
1) $7$
2) $3$
3) $5$
4) $6$

Solution:

$\mathrm{f}(\mathrm{x})=\{\begin{array}{ll}|2x^2-3x-7|& x\le -1\\ [4x^2-1]& -1<x<1\\ |x+1|+|x-2|& x⩾1\end{array}$
As the modulus function is continuous.
$\therefore \mathrm{f}(\mathrm{x})$ can be discontinuous at $\mathrm{x}=-1,1$ and at all those points where $4{\mathrm{x}}^2-1$ is integer

$\begin{array}{cccc}4x^2-1=0& ;4x^2-1=-1& ,4x^2-1=1,& 4x^2-1=2\\ x=\pm \frac{1}{2}& x=0& x=\pm \frac{1}{\sqrt{2}}& x=\pm \frac{\sqrt{3}}{2}\end{array}$
Hence discontinuous at

$\begin{array}{rl}\mathrm{x}& =\frac{1}{2},\frac{-1}{2},\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},\frac{\sqrt{3}}{2},\frac{-\sqrt{3}}{2},1\\ & =7\end{array}$
Hence, the answer is $7$.