Continuity of Composite Function

Definition of a Function

In mathematics, a function is a special type of relation that connects every element of one set to exactly one element of another set. Formally, a relation from a set $A$ to a set $B$ is said to be a function if each element of $A$ has one and only one image in $B$.

In simple terms, a function ensures that every input (from the domain) gives exactly one output (in the codomain).

If $A$ and $B$ are two non-empty sets, then a relation from $A$ to $B$ is a function if for every $x \in A$, there exists a unique $f(x) \in B$.

This is written as:
$f: A \rightarrow B$
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 Functions

Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be two given functions. The composition of functions combines these two functions into a single mapping. It is denoted by $g \circ f$ (read as “$g$ of $f$”) and is defined as the function $g \circ f : A \rightarrow C$, given by: $(g \circ f)(x) = g(f(x))$

In simple terms, the composition of functions means applying $f$ first and then $g$ to the result. This concept plays a crucial role in understanding continuity of composite functions, especially when analyzing limits and behavior of functions over an interval.

Symbol of Composition of Functions

The symbol of composition of functions is represented by the circle ($\circ$). It connects two functions such that the output of one function becomes the input of the other, allowing us to combine them into a single expression.

Representation and Reading

The composition of two functions is written as: $(f \circ g)(x) = f(g(x))$

It is read as “f of g of x.”
Here, $g(x)$ is the inner function and $f(x)$ is the outer function.

Similarly, $(g \circ f)(x) = g(f(x))$

It is read as “g of f of x.”
Here, $f(x)$ is the inner function and $g(x)$ is the outer function.

Theorem on Continuity of Composite Functions

Theorem 2: Suppose $f$ and $g$ are real-valued functions such that $(f \circ g)$ is defined at $x = c$.
If $g$ is continuous at $x = c$ and $f$ is continuous at $x = g(c)$, then $(f \circ g)$ is continuous at $x = c$.

This theorem helps determine whether a composite function is continuous at a particular point.

Example – Continuity of a Composite Function

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

Solution:
We can express $f(x)$ as a composition of two functions: $g(x) = \sin x$ and $h(x) = x^2$.
Thus, $f(x) = (g \circ h)(x) = g(h(x))$.

Since both $g(x)$ and $h(x)$ are continuous functions, by Theorem 2, the composite function $f(x)$ is continuous for all real numbers.

Checking Continuity in Composite Functions

If the function $f(x)$ is continuous at $x = a$, and another function $g(x)$ is continuous at $x = f(a)$,
then the composite function $(g \circ f)(x) = g(f(x))$ is continuous at $x = a$.

Example – When a Composite Function is Discontinuous

Consider the function $f(x) = \dfrac{1}{1 - x}$, which is discontinuous at $x = 1$.

If we define $g(x) = f(f(x))$, then $g(x)$ will not be defined whenever $f(x)$ is not defined.
Hence, $g(x)$ is also discontinuous at $x = 1$.

Now,

$g(x) = \dfrac{1}{1 - f(x)} = \dfrac{1}{1 - \dfrac{1}{1 - x}} = \dfrac{x - 1}{x}$

This shows that $g(x)$ is discontinuous at $x = 0$ and $x = 1$.

Higher-Order Composition Example

Consider,

$h(x) = f(f(f(x))) = f\left(\dfrac{x - 1}{x}\right) = \dfrac{1}{1 - \dfrac{x - 1}{x}} = x$

Although this simplifies to $h(x) = x$, the function remains discontinuous at $x = 0$ and $x = 1$,
where $f(x)$ and $f(f(x))$ respectively are not defined.

Solved Examples Based On the Continuity of Composite Functions:

Example 1: Let $f:[-1,3] \rightarrow \mathbb{R}$ be defined as

$f(x)=\left{\begin{array}{rl}
|x|+[x], & -1 \leq x < 1 \\
x+|x|, & 1 \leq x < 2 \\
x+[x], & 2 \leq x \leq 3
\end{array}\right.$

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 $f \circ g(x)$ is continuous at $x = a$ if $g$ is continuous at $x = a$ and $f$ is continuous at $g(a)$.

$f(x)=\left{\begin{array}{cc}
-x-1, & -1 \leq x < 0 \\
x+0, & 0 \leq x < 1 \\
2x, & 1 \leq x < 2 \\
x+2, & 2 \leq x < 3 \\
x+3, & x = 3
\end{array}\right.$

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

Example 2: If $f(x)=\left{\begin{array}{ll}
\sin x, & x \neq n\pi \\
2, & \text{otherwise}
\end{array}\right.$
and
$g(x)=\left{\begin{array}{ll}
x^2+1, & x \neq 0, 2 \\
4, & x=0 \\
5, & x=2
\end{array}\right.$
then the number of points in $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ where $g{f(x)}$ is discontinuous is:

1. $1$

2. $0$

3. $2$

4. None of these

Solution:

$f(x)$ is discontinuous at $x = n\pi$.
So check at $x = 0$:

$\lim_{x \rightarrow 0^{+}} g{f(x)} = \lim_{x \rightarrow 0^{+}} g{\sin x} = \lim_{x \rightarrow 0^{+}} (\sin^2 x + 1) = 1$

Similarly, $\lim_{x \rightarrow 0^{-}} g{f(x)} = 1$

But $g{f(0)} = 5$

Since $\lim_{x \rightarrow 0^{+}} g{f(x)} = \lim_{x \rightarrow 0^{-}} g{f(x)} \neq g{f(0)}$,
$g{f(x)}$ is not continuous at $x = 0$.

For all other $x$, it is continuous in the given interval.

Hence, the answer is 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:

$g(x)=x^2+5$ is continuous for all $x$.

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

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

$\Rightarrow f(g(x)) = \frac{1}{(x^2+5)-6} = \frac{1}{x^2-1}$

It is discontinuous at $x = 1$ and $x = -1$, where the denominator becomes $0$.

Hence, the answer is option (1).

Example 4: If $f(x)=\frac{1}{x-6}$, then the 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 $f(f(x))$ is discontinuous there.

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

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

$g(x)=\frac{x-6}{37-6x}$

$\therefore f(f(x))$ is not defined at $x = \frac{37}{6}$.

Hence, $f(f(x))$ is discontinuous at $x = 6$ and $x = \frac{37}{6}$.

So the number of points where it is discontinuous = 2.
Hence, the answer is option (1).

Example 5: The number of points where the function

$f(x)=\begin{cases}
|2x^2 - 3x - 7|, & x \leq -1 \\
[4x^2 - 1], & -1 < x < 1 \\
|x + 1| + |x - 2|, & x \geq 1
\end{cases}$

is discontinuous is:

1. $7$

2. $3$

3. $5$

4. $6$

Solution:

As the modulus function is continuous, $f(x)$ can be discontinuous only at $x = -1$, $x = 1$, and wherever $4x^2 - 1$ is an integer.

Now, set $4x^2 - 1 = n$, where $n$ is an integer.

For $n = 0$: $x = \pm \frac{1}{2}$
For $n = -1$: $x = 0$
For $n = 1$: $x = \pm \frac{1}{\sqrt{2}}$
For $n = 2$: $x = \pm \frac{\sqrt{3}}{2}$

Hence, the function is discontinuous at
$x = \frac{1}{2}, -\frac{1}{2}, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}, 1$

Total discontinuous points = 7

Hence, the answer is option (1).

List of topics related to Continuity of Composite Function

This section gives you a complete list of topics linked to the continuity of composite functions, including related ideas like differentiability, testing differentiability, continuity and differentiability of composite functions. It helps you build a strong base for Class 12 Maths Chapter 5 and exams like JEE Main, CUET.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Derivative as Rate Measure: Definition, Formula, Examples

Continuity And Differentiability

Differentiability of Composite Function

NCERT Resources

This section compiles all NCERT-based resources related to Chapter 5 - Continuity and Differentiability. You’ll find well-structured notes, solved examples, textbook solutions, and exemplar problems for complete preparation.

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 Continuity of Composite Function

This section includes MCQ-based practice questions designed to test your understanding of continuity in composite functions. Solving these questions will boost your speed, accuracy, and exam readiness.

Continuity Of Composite Function- Practice Question MCQ

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