It is easy to see that $f$ is one-one and onto, so that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}=\{(1,3),(3,2),(2,1)\}=f$
It is easy to see that $f$ is one-one and onto, so that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}=\{(1,3),(3,2),(2,1)\}=f$
It is easy to see that $f$ is one-one and onto, so that $f$ is invertible with $f^{-1}=\{(3,1),(2,3),(1,2)\}$
Similar Questions
Consider $f:\{1,2,3\} \rightarrow\{a, b, c\}$ given by $f(1)=a, \,f(2)=b$ and $f(3)=c .$ Find $f^{-1}$ and show that $\left(f^{-1}\right)^{-1}=f$.
Consider $f:\{1,2,3\} \rightarrow\{a, b, c\}$ given by $f(1)=a, \,f(2)=b$ and $f(3)=c .$ Find $f^{-1}$ and show that $\left(f^{-1}\right)^{-1}=f$.
If $X$ and $Y$ are two non- empty sets where $f:X \to Y$ is function is defined such that $f(c) = \left\{ {f(x):x \in C} \right\}$ for $C \subseteq X$ and ${f^{ - 1}}(D) = \{ x:f(x) \in D\} $ for $D \subseteq Y$ for any $A \subseteq X$ and $B \subseteq Y,$ then
If $X$ and $Y$ are two non- empty sets where $f:X \to Y$ is function is defined such that $f(c) = \left\{ {f(x):x \in C} \right\}$ for $C \subseteq X$ and ${f^{ - 1}}(D) = \{ x:f(x) \in D\} $ for $D \subseteq Y$ for any $A \subseteq X$ and $B \subseteq Y,$ then
- [IIT 2005]
Let the function $f$ be defined by $f(x) = \frac{{2x + 1}}{{1 - 3x}}$, then ${f^{ - 1}}(x)$ is
Let the function $f$ be defined by $f(x) = \frac{{2x + 1}}{{1 - 3x}}$, then ${f^{ - 1}}(x)$ is
Let $S=\{a, b, c\}$ and $T=\{1,2,3\} .$ Find $F^{-1}$ of the following functions $F$ from $S$ to $T$. if it exists. $F =\{( a , 2)\,,(b , 1),\,( c , 1)\}$
Let $S=\{a, b, c\}$ and $T=\{1,2,3\} .$ Find $F^{-1}$ of the following functions $F$ from $S$ to $T$. if it exists. $F =\{( a , 2)\,,(b , 1),\,( c , 1)\}$
If $y = f(x) = \frac{{x + 2}}{{x - 1}}$, then $x = $
If $y = f(x) = \frac{{x + 2}}{{x - 1}}$, then $x = $
- [IIT 1984]