# Consider $f:\left\{ 1,2,3 \right\}\to \left\{ a,b,c \right\}$ and $g:\left\{ a,b,c \right\}\to \left\{ apple,ball,cat \right\}$ defined as $f(1)=a,f(2)=b,f(3)=c,g(a)=apple,g(b)=ball,g(c)=cat$ . Show that $f,g,g\circ f$ are invertible. Find ${{f}^{-1}},{{g}^{-1}}$ and ${{\left( g\circ f \right)}^{-1}}$ . Show that ${{\left( g\circ f \right)}^{-1}}={{f}^{-1}}\circ {{g}^{-1}}$ .

Answer

Verified

383.4k+ views

Hint: At first we have to check if the functions $f,g$ are one to one and onto or not. A function is invertible only if the function is both one to one and onto.

Complete step-by-step answer:

The function f is defined as:

$f:\left\{ 1,2,3 \right\}\to \left\{ a,b,c \right\}$.

$f(1)=a,f(2)=b,f(3)=c$ .

Here $\left\{ 1,2,3 \right\}$ is the domain of the function f. $\left\{ a,b,c \right\}$ is the codomain of the function f.

We know that a function is said to be one to one if every different element of the domain has different images.

Here image of 1 is a. Image of 2 is b. Image of 3 is c. Therefore, every different element of the domain has a different image. Hence, f is one to one.

We know that a function is said to be onto if for every element of the codomain, we can find out at least one preimage from the domain.

The preimage of a is 1. Preimage of b is 2. Preimage of c is 3.

Therefore, every element of the codomain has a preimage. Hence, f is a onto function.

Therefore, f is both one to one and onto. So, f is invertible.

Similarly, the function g is defined as:

$g:\left\{ a,b,c \right\}\to \left\{ apple,ball,cat \right\}$

$g(a)=apple, g(b)=ball, g(c)=cat$

Here $\left\{ a,b,c \right\}$ is the domain of the function. $\left\{ apple,ball,cat \right\}$ is the codomain of the function.

The function g is one to one as image of a is apple, image of b is ball, image of c is cat. Therefore every element of the domain has a different image.

The function g is onto as preimage of apple is a, preimage of ball is b, preimage of cat is c. Therefore every element of the codomain has one preimage in the domain.

Hence, g is both one to one and onto. So, g is invertible.

Now, $(g\circ f):\left\{ 1,2,3 \right\}\to \left\{ apple,ball,cat \right\}$ is defined as:

As $\left( g\circ f \right)\left( 1 \right)=g\left( f\left( 1 \right) \right)=g\left( a \right)=apple$

$\begin{align}

& (g\circ f)(2)=g\left( f\left( 2 \right) \right)=g(b)=ball \\

& (g\circ f)(3)=g\left( f\left( 3 \right) \right)=g(c)=cat \\

\end{align}$

$g\circ f$ is one to one as every element of the domain $\left\{ 1,2,3 \right\}$ has different image.

$g\circ f$ is onto as every element of the codomain $\left\{ apple,ball,cat \right\}$ has a preimage in the domain.

Therefore $g\circ f$ is invertible.

We know that if a function $f$ maps one element $x$ to $y$, then the inverse function maps the image $y$ to $x$. That is:

$f\left( x \right)=y\Rightarrow {{f}^{-1}}\left( y \right)=x$

Therefore,

$\begin{align}

& f(1)=a\Rightarrow {{f}^{-1}}\left( a \right)=1 \\

& f(2)=b\Rightarrow {{f}^{-1}}\left( b \right)=2 \\

& f(3)=c\Rightarrow {{f}^{-1}}\left( c \right)=3 \\

\end{align}$

Hence,

${{f}^{-1}}:\left\{ a,b,c \right\}\to \left\{ 1,2,3 \right\}$ , such that:

${{f}^{-1}}\left( a \right)=1,{{f}^{-1}}\left( b \right)=2,{{f}^{-1}}\left( c \right)=3$

Similarly,

$\begin{align}

&g\left( a \right)=apple\Rightarrow {{g}^{-1}}\left( apple \right)=a \\

&g\left( b \right)=ball\Rightarrow {{g}^{-1}}\left( ball \right)=b \\

&g\left( c \right)=cat\Rightarrow {{g}^{-1}}\left( cat \right)=c \\

\end{align}$

Therefore,

${{g}^{-1}}:\left\{ apple,ball,cat \right\}\to \left\{ a,b,c \right\}$ , such that:

$\begin{align}

& {{g}^{-1}}(apple)=a \\

& {{g}^{-1}}\left( ball \right)=b \\

& {{g}^{-1}}\left( cat \right)=c \\

\end{align}$

Similarly,

$\begin{align}

& \left( g\circ f \right)\left( 1 \right)=apple\Rightarrow {{\left( g\circ f \right)}^{-1}}\left( apple \right)=1 \\

& \left( g\circ f \right)\left( 2 \right)=ball\Rightarrow {{\left( g\circ f \right)}^{-1}}\left( ball \right)=2 \\

& \left( g\circ f \right)\left( 3 \right)=cat\Rightarrow {{\left( g\circ f \right)}^{-1}}\left( cat \right)=3 \\

\end{align}$

Therefore,

${{\left( g\circ f \right)}^{-1}}:\left\{ apple,ball,cat \right\}\to \left\{ 1,2,3 \right\}$ , such that:

$\begin{align}

& {{\left( g\circ f \right)}^{-1}}\left( apple \right)=1 \\

& {{\left( g\circ f \right)}^{-1}}\left( ball \right)=b \\

& {{\left( g\circ f \right)}^{-1}}\left( cat \right)=c \\

\end{align}$

Now,

$\begin{align}

& \left( {{f}^{-1}}\circ {{g}^{-1}} \right)\left( apple \right)={{f}^{-1}}\left( {{g}^{-1}}\left( apple \right) \right)={{f}^{-1}}\left( a \right)=1 \\

& \left( {{f}^{-1}}\circ {{g}^{-1}} \right)\left( ball \right)={{f}^{-1}}\left( {{g}^{-1}}\left( ball \right) \right)={{f}^{-1}}\left( b \right)=2 \\

& \left( {{f}^{-1}}\circ {{g}^{-1}} \right)\left( cat \right)={{f}^{-1}}\left( {{g}^{-1}}\left( cat \right) \right)={{f}^{-1}}\left( c \right)=3 \\

& \\

\end{align}$

Therefore,

$\begin{align}

& {{\left( g\circ f \right)}^{-1}}\left( apple \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( apple \right) \\

& {{\left( g\circ f \right)}^{-1}}\left( ball \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( ball \right) \\

& {{\left( g\circ f \right)}^{-1}}\left( cat \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( cat \right) \\

\end{align}$

Hence, ${{\left( g\circ f \right)}^{-1}}=\left( {{f}^{-1}}\circ {{g}^{-1}} \right)$

Note: We generally make mistakes to find out the inverse function. Always remember:

$f\left( x \right)=y\Rightarrow {{f}^{-1}}\left( y \right)=x$

Complete step-by-step answer:

The function f is defined as:

$f:\left\{ 1,2,3 \right\}\to \left\{ a,b,c \right\}$.

$f(1)=a,f(2)=b,f(3)=c$ .

Here $\left\{ 1,2,3 \right\}$ is the domain of the function f. $\left\{ a,b,c \right\}$ is the codomain of the function f.

We know that a function is said to be one to one if every different element of the domain has different images.

Here image of 1 is a. Image of 2 is b. Image of 3 is c. Therefore, every different element of the domain has a different image. Hence, f is one to one.

We know that a function is said to be onto if for every element of the codomain, we can find out at least one preimage from the domain.

The preimage of a is 1. Preimage of b is 2. Preimage of c is 3.

Therefore, every element of the codomain has a preimage. Hence, f is a onto function.

Therefore, f is both one to one and onto. So, f is invertible.

Similarly, the function g is defined as:

$g:\left\{ a,b,c \right\}\to \left\{ apple,ball,cat \right\}$

$g(a)=apple, g(b)=ball, g(c)=cat$

Here $\left\{ a,b,c \right\}$ is the domain of the function. $\left\{ apple,ball,cat \right\}$ is the codomain of the function.

The function g is one to one as image of a is apple, image of b is ball, image of c is cat. Therefore every element of the domain has a different image.

The function g is onto as preimage of apple is a, preimage of ball is b, preimage of cat is c. Therefore every element of the codomain has one preimage in the domain.

Hence, g is both one to one and onto. So, g is invertible.

Now, $(g\circ f):\left\{ 1,2,3 \right\}\to \left\{ apple,ball,cat \right\}$ is defined as:

As $\left( g\circ f \right)\left( 1 \right)=g\left( f\left( 1 \right) \right)=g\left( a \right)=apple$

$\begin{align}

& (g\circ f)(2)=g\left( f\left( 2 \right) \right)=g(b)=ball \\

& (g\circ f)(3)=g\left( f\left( 3 \right) \right)=g(c)=cat \\

\end{align}$

$g\circ f$ is one to one as every element of the domain $\left\{ 1,2,3 \right\}$ has different image.

$g\circ f$ is onto as every element of the codomain $\left\{ apple,ball,cat \right\}$ has a preimage in the domain.

Therefore $g\circ f$ is invertible.

We know that if a function $f$ maps one element $x$ to $y$, then the inverse function maps the image $y$ to $x$. That is:

$f\left( x \right)=y\Rightarrow {{f}^{-1}}\left( y \right)=x$

Therefore,

$\begin{align}

& f(1)=a\Rightarrow {{f}^{-1}}\left( a \right)=1 \\

& f(2)=b\Rightarrow {{f}^{-1}}\left( b \right)=2 \\

& f(3)=c\Rightarrow {{f}^{-1}}\left( c \right)=3 \\

\end{align}$

Hence,

${{f}^{-1}}:\left\{ a,b,c \right\}\to \left\{ 1,2,3 \right\}$ , such that:

${{f}^{-1}}\left( a \right)=1,{{f}^{-1}}\left( b \right)=2,{{f}^{-1}}\left( c \right)=3$

Similarly,

$\begin{align}

&g\left( a \right)=apple\Rightarrow {{g}^{-1}}\left( apple \right)=a \\

&g\left( b \right)=ball\Rightarrow {{g}^{-1}}\left( ball \right)=b \\

&g\left( c \right)=cat\Rightarrow {{g}^{-1}}\left( cat \right)=c \\

\end{align}$

Therefore,

${{g}^{-1}}:\left\{ apple,ball,cat \right\}\to \left\{ a,b,c \right\}$ , such that:

$\begin{align}

& {{g}^{-1}}(apple)=a \\

& {{g}^{-1}}\left( ball \right)=b \\

& {{g}^{-1}}\left( cat \right)=c \\

\end{align}$

Similarly,

$\begin{align}

& \left( g\circ f \right)\left( 1 \right)=apple\Rightarrow {{\left( g\circ f \right)}^{-1}}\left( apple \right)=1 \\

& \left( g\circ f \right)\left( 2 \right)=ball\Rightarrow {{\left( g\circ f \right)}^{-1}}\left( ball \right)=2 \\

& \left( g\circ f \right)\left( 3 \right)=cat\Rightarrow {{\left( g\circ f \right)}^{-1}}\left( cat \right)=3 \\

\end{align}$

Therefore,

${{\left( g\circ f \right)}^{-1}}:\left\{ apple,ball,cat \right\}\to \left\{ 1,2,3 \right\}$ , such that:

$\begin{align}

& {{\left( g\circ f \right)}^{-1}}\left( apple \right)=1 \\

& {{\left( g\circ f \right)}^{-1}}\left( ball \right)=b \\

& {{\left( g\circ f \right)}^{-1}}\left( cat \right)=c \\

\end{align}$

Now,

$\begin{align}

& \left( {{f}^{-1}}\circ {{g}^{-1}} \right)\left( apple \right)={{f}^{-1}}\left( {{g}^{-1}}\left( apple \right) \right)={{f}^{-1}}\left( a \right)=1 \\

& \left( {{f}^{-1}}\circ {{g}^{-1}} \right)\left( ball \right)={{f}^{-1}}\left( {{g}^{-1}}\left( ball \right) \right)={{f}^{-1}}\left( b \right)=2 \\

& \left( {{f}^{-1}}\circ {{g}^{-1}} \right)\left( cat \right)={{f}^{-1}}\left( {{g}^{-1}}\left( cat \right) \right)={{f}^{-1}}\left( c \right)=3 \\

& \\

\end{align}$

Therefore,

$\begin{align}

& {{\left( g\circ f \right)}^{-1}}\left( apple \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( apple \right) \\

& {{\left( g\circ f \right)}^{-1}}\left( ball \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( ball \right) \\

& {{\left( g\circ f \right)}^{-1}}\left( cat \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( cat \right) \\

\end{align}$

Hence, ${{\left( g\circ f \right)}^{-1}}=\left( {{f}^{-1}}\circ {{g}^{-1}} \right)$

Note: We generally make mistakes to find out the inverse function. Always remember:

$f\left( x \right)=y\Rightarrow {{f}^{-1}}\left( y \right)=x$

Recently Updated Pages

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is 1 divided by 0 class 8 maths CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Convert compound sentence to simple sentence He is class 10 english CBSE

India lies between latitudes and longitudes class 12 social science CBSE

Why are rivers important for the countrys economy class 12 social science CBSE

Distinguish between Khadar and Bhangar class 9 social science CBSE