Continuity of Composite Function

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
$f: A \rightarrow B$ 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 function

Let $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ and $\mathrm{g}: \mathrm{B} \rightarrow \mathrm{C}$ be two functions. Then the composition of f and $g$ is denoted by gof and defined as the function gof : $A \rightarrow C$ 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 ( $f o$ $\mathrm{g})$ 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 $g(x), \quad g(x)=\frac{1}{1-f(x)}=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$
It is discontinuous at $x=0$
So, $g(x)=f(f(x))$ is discontinuous at $x=0$ and $x=1$

Now consider,

$
\begin{aligned}
\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{aligned}
$

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.

Recommended Video Based on Continuity of Composite Functions:

Solved Examples Based On the Continuity of Composite Functions:

Example 1: Let $f:[-1,3] \rightarrow R{\text {be defined as }}$ $f(x)=\left\{\begin{array}{rlrl}|x|+[x], & & -1 & \leq x<1 \\ x+|x|, & & 1 & \leq x<2 \\ x+[x], & & 2 \leq x \leq 3\end{array}\right.$ 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 fog $(x)$ is continuous at $x=a$ if $g$ is continuous at $x=$ a and f is continuous at $\mathrm{g}(\mathrm{a})$.

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

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

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

similarly $\lim\limits_{x \rightarrow 0^{-}} g\{f(x)\}=1 g\{f(0)\}=5 \lim\limits_{x \rightarrow 0^{+}} g\{f(x)\}=\lim\limits_{x \rightarrow 0^{-}} g\{f(x)\} \neq g\{f(0)\} g\{f(x)\}$ 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}{\left(x^2+5\right)-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{aligned}
& 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}{37-6 x}
\end{aligned}
$

$\mathrm{f}(\mathrm{f}(\mathrm{x}))$ 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
$\mathrm{f}(\mathrm{x})= \begin{cases}\left|2 x^2-3 x-7\right| & \text { if } x \leq-1 \\ {\left[4 x^2-1\right]} & \text { if }-1<x<1 \\ |x+1|+|x-2| & \text { if } x \geqslant 1, \quad[\mathrm{t}]_{\text {denotes the greatest integer }} \leqslant \mathrm{t} \text {, is discontinuous is }\end{cases}$ $\qquad$
1) $7$
2) $3$
3) $5$
4) $6$

Solution:

$
\mathrm{f}(\mathrm{x})= \begin{cases}\left|2 x^2-3 x-7\right| & x \leq-1 \\ {\left[4 x^2-1\right]} & -1<x<1 \\ |x+1|+|x-2| & x \geqslant 1\end{cases}
$
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}
4 x^2-1=0 & ; 4 x^2-1=-1 & , 4 x^2-1=1, & 4 x^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{aligned}
\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{aligned}
$
Hence, the answer is $7$.